 Solution

For a nonrotating inviscid Boussinesq fluid, start from $\nabla \cdot \mathbf{u}=0$, $D\rho /Dt=0$ andThe resting background has ${\bar{p}}_z=-g\bar{\rho }$ and $N^2=-g{\bar{\rho }}_z/{\rho }_0>0$. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$, $p=\bar{p}+p^{\prime }$ and $\sigma =-g{\rho }^{\prime }/{\rho }_0$. Keeping first-order terms givesThe PDF has a genuine sign error: its momentum equation uses a minus sign before $\sigma \hat{\mathbf{z}}$ despite defining $\sigma =-g{\rho }^{\prime }/{\rho }_0$. A lighter parcel has $\sigma >0$ and accelerates upward, so the plus sign is necessary. Retaining both printed signs would yield negative squared wave frequency and exponential growth in a supposedly stable fluid; no oscillatory stable-wave dispersion relation follows from that inconsistent system. Alternatively one can define a downward density variable $\tilde{\sigma }=g{\rho }^{\prime }/{\rho }_0$, but then its density equation must have ${\tilde{\sigma }}_t=+N^2{w}^{\prime }$.

For the consistent system, take a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, with $k\ne 0$. Incompressibility gives $ku+mw=0$. Project the momentum equation perpendicular to the wavevector, eliminating pressure: $-i\omega w=k^2\sigma /(k^2+m^2)$. Combining with $-i\omega \sigma =-N^{2}w$ givesOn a smooth signed-frequency branch, the phase velocity and group velocity areTheir scalar product is zero. Equivalently, the dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler homogeneous function theorem gives $(k,m)\cdot {\nabla }_{(k,m)}\omega =0$. The phase velocity moves crests normal to their phase lines; the group velocity moves wave packets and energy along those lines. Their perpendicularity explains why following individual crests does not reveal the direction of energy propagation.