= Solution
At $O(\epsilon\gamma)$, the slowly varying <wave envelope> obeys
$$
\boxed{
\partial_{\widehat t}\widetilde\psi
+\mathbf c_g\mathbin{\cdot}
\widehat\nabla\widetilde\psi=0
},
$$
where the <group velocity> obtained from $\omega=Nk/(k^2+m^2)^{1/2}$ is
$$
\boxed{
\mathbf c_g
=\left(
\frac{Nm^2}{(k^2+m^2)^{3/2}},
0,
-\frac{Nkm}{(k^2+m^2)^{3/2}}
\right)
}.
$$
It satisfies $\mathbf k\cdot\mathbf c_g=0$. If initially $\mathbf k\cdot\widehat\nabla\widetilde\psi=0$, the envelope structure varies across the wavevector and translates unchanged in the direction $\mathbf c_g$ at speed
$$
|\mathbf c_g|=\frac{N|m|}{k^2+m^2}.
$$
\b[Thus 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.]
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