Phase and group velocity calculation

Question

I have a wave described by $A \sin(k_x x) \exp(i(k_y y− ωt))$. This wave is propagating in $+y$ direction. However its angular frequency is given by $k_x^2 + k_y^2 = (w/c)^2$. If I want to calculate phase and group velocity does it mean that I only use $k_y$, because the wave is travelling in that direction? For example $v_{\rm p} = \omega/k_y$ and $v_{\rm g} = d\omega/dk_y$?

Answer 1

The phase velocity is the easiest place to start. Ask yourself what is the definition of a wavefront, and then how fast the wavefronts are moving.

Your wave has the form $$ f(x) e^{i(k_y y - \omega t)} $$ where $f(x)$ is some function of $x$; in this case $A \sin(k_x x)$. So you have a wave travelling in $+y$ direction as you say, and all that $f(x)$ is doing is saying that the amplitude of the wave is a function of $x$. I hope this is enough to make it clear to you that the phase velocity is indeed $v_{\rm p} = \omega/k_y$ since this is the speed at which the wavefronts are moving. (A wavefront being a locus of points at constant phase of the wave).

Once we have the phase velocity at each $\omega$ and $k_y$, the group velocity follows. To see this, consider a pair of waves of the type under discussion, one at $k_x, k_y$ and one at $k_x + dk_x, \; k_y + dk_y$. Add the two waves together, and you should find you have a fast oscillation and a beat pattern envelope. That envelope moves at speed $d \omega / d k_y$. Using the dispersion relation you then get $$ \frac{d \omega}{d k_y} = c \frac{d}{d k_y} \left(k_x^2 + k_y^2 \right)^{1/2} $$ and to carry out the differentiation you would need to know how $k_x$ and $k_y$ are related for the waves you are considering.