Group Velocity of EMW: Conditions \& Expression

[dingo_d]

Homework Statement

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

Homework Equations

The expression for group velocity of the traveling wave is $v_g=\frac{d\omega}{dk}$
where $\omega$ is angular frequency and k is wave number (vector in 3D).

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 anyone know how to start, and what to look? Thnx!

 

[jdwood983]

You (should) know that the phase velocity is also found from

$$v_p=c\cdot n(\omega)$$

Do you know of any relations between the dielectric permittivity you are given and the real refractive index $n(\omega)$?

 

[dingo_d]

We have learned that $v_p=\frac{c}{n(k)}$, and our substitute assistant said sth about:

$k^2=\mu\epsilon\frac{\omega^2}{c^2}$

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)

 

[gabbagabbahey]

$k^2=\mu\epsilon\frac{\omega^2}{c^2}$

This is all you need to solve this problem...For most media, $\mu\approx\mu_0$, so I think that is a fair assumption to make here (Or at least assume that $\mu$ is independent of $\omega$). After that, just substitute in your expression for $\epsilon$ and find $\frac{d\omega}{d k}$ 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 $k$ and angular frequency $\omega$) and seeing what conditions are necessary for such fields to satisfy Maxwell's equations (or, equivalently, the wave equation) in a medium with permittivity $\epsilon$.

 

[jdwood983]

Whoops, I was thinking phase velocity and you wanted group velocity...

 

[dingo_d]

This is all you need to solve this problem...For most media, $\mu\approx\mu_0$, so I think that is a fair assumption to make here (Or at least assume that $\mu$ is independent of $\omega$). After that, just substitute in your expression for $\epsilon$ and find $\frac{d\omega}{d k}$ 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 $k$ and angular frequency $\omega$) and seeing what conditions are necessary for such fields to satisfy Maxwell's equations (or, equivalently, the wave equation) in a medium with permittivity $\epsilon$.

Thank you very much! I'll do that! :)

 

[gabbagabbahey]

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

$$k^2=\mu\epsilon \omega^2=\mu_r\epsilon_r \frac{\omega^2}{c^2}$$

(And of course, this is only true for a non-conducting, source-free medium)

 

[dingo_d]

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

$$k^2=\mu\epsilon \omega^2=\mu_r\epsilon_r \frac{\omega^2}{c^2}$$

(And of course, this is only true for a non-conducting, source-free medium)

Yeah, because $\mu =\mu_0\mu_r$? Right? And $\mu_0\epsilon_0=\frac{1}{c^2}$...

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

$$v_g=\frac{d\omega}{dk}=\frac{1}{2\omega}\frac{d\omega^2}{dk}$$?

 

[gabbagabbahey]

Yeah, because $\mu =\mu_0\mu_r$? Right? And $\mu_0\epsilon_0=\frac{1}{c^2}$...

Yup.

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

$$v_g=\frac{d\omega}{dk}=\frac{1}{2\omega}\frac{d\omega^2}{dk}$$?

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

$$k^2=f(\omega)\implies 2k=f'(\omega)\frac{d\omega}{d k}\implies \frac{d\omega}{d k}=2\frac{\sqrt{f(\omega)}}{f'(\omega)}$$

 

[dingo_d]

... I suggest an easier method is as follows:

$$k^2=f(\omega)\implies 2k=f'(\omega)\frac{d\omega}{d k}\implies \frac{d\omega}{d k}=2\frac{\sqrt{f(\omega)}}{f'(\omega)}$$

Oh, ok! :D So I put my expression for $$\epsilon$$ in the $$k^2$$ expression and derive it, and I get my $$v_g=\frac{d\omega}{dk}$$ 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:

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

What conditions should $$\omega_p$$ and $$\omega_0$$ satisfy so that my group velocity is smaller or equal than c? Do I solve the inequality?