# Group Velocity of EMW: Conditions \& Expression

• dingo_d
In summary: What conditions should \omega_p and \omega_0 satisfy so that my group velocity is smaller or equal than c? Do I solve the...In summary, the relativity principle states that the group velocity of an electromagnetic wave traveling through a medium is smaller than the speed of light in that medium.
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!

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 $n(\omega)$?

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)

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

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 $\epsilon$.

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

gabbagabbahey said:
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! :)

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)

gabbagabbahey said:
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?

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

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dingo_d said:
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:

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

gabbagabbahey said:
... 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?

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## 1. What is group velocity?

Group velocity refers to the speed at which the energy of a wave packet (a group of waves) propagates through a medium. It is a property of a wave that describes the movement of the entire group of waves, rather than just an individual wave.

## 2. How is group velocity different from phase velocity?

Phase velocity refers to the speed at which an individual wave within a wave packet travels. Group velocity takes into account the combined effect of all the waves in the packet and describes the overall speed at which the packet moves.

## 3. What conditions affect the group velocity of electromagnetic waves?

The group velocity of electromagnetic waves is affected by the properties of the medium through which they are propagating, such as its refractive index and dispersion. It is also influenced by the frequency and direction of the waves.

## 4. What is the expression for calculating group velocity?

The expression for group velocity can vary depending on the specific type of wave and the properties of the medium. In general, it is calculated as the derivative of the wave's frequency with respect to its wavenumber.

## 5. How is group velocity used in practical applications?

In practical applications, group velocity is important in understanding the behavior of electromagnetic waves and their interactions with different materials. It is used in fields such as optics, telecommunications, and radar technology.

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