Internal Wave

Phase speed is perpendicular to group speed

Let us assume that a fluid is inviscid, linear, non-diffusive, Boussinesq, and consists of motion independent of the $y$ coordinate. The Navier-Stokes equations can therefore be written using the stream-function, $\psi$, as

$(\nabla ^{2}\psi )_{tt}=-N^{2}(z)\psi _{xx}$.

Making the wave ansatz, $\psi =\exp {\left({\vec {k}}\cdot {\vec {x}}-\omega t\right)}$, where ${\vec {k}}=(k_{x},k_{z})$ and ${\vec {x}}=(x,z)$, the dispersion relation is

Failed to parse (unknown function "\lvert"): {\displaystyle \omega = \frac{N k_x}{\lvert\vec{k}\rvert} } .

The phase speed is thus

Failed to parse (unknown function "\lvert"): {\displaystyle c_p = \frac{\omega}{\lvert\vec{k}\rvert} \hat{k} = \frac{\omega}{\lvert\vec{k}\rvert^2} \vec{k} = \frac{Nk_x}{\lvert\vec{k}\rvert^3}\vec{k} = \frac{Nk_x}{\lvert\vec{k}\rvert^3}\left( k_x, k_z \right)} .

The group speed is

Failed to parse (unknown function "\lvert"): {\displaystyle \begin{align} c_g &= \left( \frac{\partial\omega}{\partial k_x}, \frac{\partial\omega}{\partial k_z} \right)\\ &= \left( \frac{N}{\lvert\vec{k}\rvert} - \frac{N k_x^2}{\lvert\vec{k}\rvert^3}, - \frac{N k_x k_z}{\lvert\vec{k}\rvert^3} \right)\\ &= \left( \frac{N\left( k_x^2 + k_z^2\right)}{\lvert\vec{k}\rvert^3} - \frac{N k_x^2}{\lvert\vec{k}\rvert^3}, - \frac{N k_x k_z}{\lvert\vec{k}\rvert^3} \right)\\ &= \frac{N k_z}{\lvert\vec{k}\rvert^3}\left( k_z, - k_x \right)\\ \end{align}}

Therefore we find that the group velocity is perpendicular to the phase velocity!

Failed to parse (unknown function "\lvert"): {\displaystyle c_p \cdot c_g = \frac{N^2 k_x k_z}{\lvert\vec{k}\rvert^6}\left( k_x, k_z \right) \cdot \left( k_z, -k_z \right) = 0}