# Phase and Group Velocity of Electromagnetic Waves

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

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.