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Moving charges produce oscillating electric and magnetic fields -we have an electromagnetic wave.

  1. In terms of moving charges or at the level of charges, what is phase velocity and group velocity of an electromagnetic wave?

  2. What is the origin of these velocities?

I am to not able to relate the definitions of phase velocity(how fast the phase of a wave is moving) and group velocity(velocity of the envelope) to moving charges.

I am trying to get an intuitive picture of this. Please help me.

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Are you thinking of electromagnetic waves propagating in vacuum or in a medium? With respect to the group/phase velocity distinction, the two cases are very, very different. –  Emilio Pisanty Nov 22 '12 at 17:03
I was thinking in terms of vacuum. I would also like to know the distinction in a medium. –  Spaceman Spiff Nov 22 '12 at 17:28

1 Answer 1

Well, in vacuum the group velocity and phase velocity are identical. The velocity of light in vacuum is not really determined by the properties of the moving charges. It is a fundamental constant of nature: if you like, it is determined by the properties of the vacuum. The important feature of the charges that generate the EM wave is the frequency at which they oscillate, which gives the frequency of the resulting wave.

The distinction between phase and group velocity becomes important when you are in a dispersive medium, meaning a medium in which you have charges that are stuck in one place somehow, e.g. electrons bound to atomic nuclei. (An example of a dispersive medium is a piece of glass.) Now the speed at which light propagates will depend on its frequency, this is what dispersion means.

Basically, dispersion arises because the bound charges have their own frequency that they like to oscillate at. However, the incoming light forces them to oscillate at a different frequency. The way they react to the incoming light (the amplitude of their oscillation) depends on how the light's frequency compares to their own favourite frequency. The bound charges' reaction is really important because the total EM wave that you observe is the sum of the incoming wave and the contribution from the oscillating charges, which re-radiate EM waves at an amplitude depending on their amplitude of oscillation.

Given that different frequencies of light have different speeds (i.e. phase velocities), we can now see why the group velocity is different, in general. The phase velocity is $$ v_p = \frac{\omega}{k}, $$ and the group velocity is $$ v_g = \frac{\partial \omega}{\partial k}, $$ which are only the same if $\omega = c k$, with $c$ a constant (the speed of light). This is why $v_p = v_g$ in vacuum. (Here $\omega$ is angular frequency and $k = 2\pi/\lambda$ is wavenumber.) However, if different frequencies have different velocities, then $v_p(\omega)$ is a non-trivial function of $\omega$, which is only true if the relationship between $\omega$ and $k$ (the dispersion relation) is more complicated than just the linear one. In other words, in a dispersive medium, $\omega \neq c k$. But the condition $v_p = v_g$ implies that $$\frac{\partial \omega}{\partial k} = \frac{\omega}{k} \; \Leftrightarrow \; \omega = c k, $$ which shows that if dispersion is present then $v_p \neq v_g$. To see why group velocity is a useful measure of how a wavepacket moves, which is the usual interpretation, see e.g. Wikipedia.

The microscopic origin of dispersion is non-trivial, so I'll leave it to greater luminaries than I to give a proper explanation. This was really just a comment that got a bit long. Good question though, hopefully someone really clever will go into all the gory details :)

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