 Solution

At $O(ϵ\gamma )$, the slowly varying wave envelope obeyswhere the group velocity obtained from $\omega =Nk/(k^2+m^2{)}^{1/2}$ isIt satisfies $\mathbf{k}\cdot {\mathbf{c}}_g=0$. If initially $\mathbf{k}\cdot \hat{\nabla }\tilde{\psi }=0$, the envelope structure varies across the wavevector and translates unchanged in the direction ${\mathbf{c}}_g$ at speedThus internal-wave crests propagate with phase velocity parallel to $\mathbf{k}$, while the packet and its energy propagate perpendicular to $\mathbf{k}$ with the group velocity.