# Determine Phase and Group Velocities for Monoatomic Lattice

I have a question about determining the phase and group velocity for a monoatomic lattice.

I know from various reference texts that

$$v_p = \frac {\omega}{q}$$ $$v_g = \frac {\partial \omega}{\partial q}$$

Where q is the wave vector and $$\omega$$ is the angular frequency.

My question is, why? This seems to be derived out of thin air, and I'm not sure what the physical/mathematical explanation for this is.

Thank you,

So the derivation of the first is just that you look at a plane wave $$\exp(i (qx-\omega(q) t))$$ and the speed that it is going is just $$\omega(q)/q.$$
The derivation of the second is more subtle, you consider a Gaussian wave packet in the $$q$$-space $$f(x,t)=\int\mathrm dq~e^{-aq^2}e^{i (qx-\omega(q) t)}$$ where this is a valid solution of the wave equation due to it being a linear combination of valid solutions to the wave equation. Now Taylor expand $$\omega(q)$$ out to second order to find a Gaussian which is moving at speed $$\omega'(q)$$ and maybe also diffusing with a rate that has something to do with $$\omega''(q).$$ Inside it still has the same phase factor $$\omega/q$$ but the Gaussian envelope itself is moving at a different speed than the wave it contains inside of it.
Another derivation, as Pieter mentions in comments below, is to just look at a sum $$e^{i(qx-\omega(q) t)} + e^{i((q+\delta q)x-\omega(q+\delta q) t)} =e^{i(qx-\omega(q) t)} (1 + e^{i \delta q(x - \omega'(q) t)})$$ This thing on the right can be interpreted as a sort of envelope in which the wave on the left lives, and it has this pattern of $$f(x-vt)$$, a function traveling forward with speed $$v$$, for $$v=\omega'(q).$$