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

$\omega ={\frac {Nk_{x}}{\left|{\vec {k}}\right|}}$

where $\left|{\vec {k}}\right|={\sqrt {k_{x}^{2}+k_{z}^{2}}}$.

The phase speed is thus

${\vec {c}}_{p}={\frac {\omega }{\left|{\vec {k}}\right|}}{\hat {k}}={\frac {\omega }{\left|{\vec {k}}\right|^{2}}}{\vec {k}}={\frac {Nk_{x}}{\left|{\vec {k}}\right|^{3}}}{\vec {k}}={\frac {Nk_{x}}{\left|{\vec {k}}\right|^{3}}}\left(k_{x},k_{z}\right)$.

The group speed is

${\begin{aligned}{\vec {c}}_{g}&=\left({\frac {\partial \omega }{\partial k_{x}}},{\frac {\partial \omega }{\partial k_{z}}}\right)\\&=\left({\frac {N}{\left|{\vec {k}}\right|}}-{\frac {Nk_{x}^{2}}{\left|{\vec {k}}\right|^{3}}},-{\frac {Nk_{x}k_{z}}{\left|{\vec {k}}\right|^{3}}}\right)\\&=\left({\frac {N\left(k_{x}^{2}+k_{z}^{2}\right)}{\left|{\vec {k}}\right|^{3}}}-{\frac {Nk_{x}^{2}}{\left|{\vec {k}}\right|^{3}}},-{\frac {Nk_{x}k_{z}}{\left|{\vec {k}}\right|^{3}}}\right)\\&={\frac {Nk_{z}}{\left|{\vec {k}}\right|^{3}}}\left(k_{z},-k_{x}\right)\\\end{aligned}}$

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

${\vec {c}}_{p}\cdot {\vec {c}}_{g}={\frac {N^{2}k_{x}k_{z}}{\left|{\vec {k}}\right|^{6}}}\left(k_{x},k_{z}\right)\cdot \left(k_{z},-k_{x}\right)=0$

This is completely counterintuitive from our experiences with surface waves, sound waves, and in fact most other waves, where the energy propagates in the same direction as the phase. The Saint Andrew's Cross experiment is the classic example demonstrating this. See Kundu for pictures and further elaboration.

Waves in a channel

Consider two-dimensional flow between boundaries at $z=0$ and $z=H$. The boundary conditions are $w=0$ at $z=0,H$. Since $w=-\psi _{x}$ by the definition of the stream-function, the boundary condition becomes $\psi _{x}=0{\text{ at }}z=0,H$. The equation of motion is

$(\nabla ^{2}\psi )_{tt}=-N^{2}(z)\psi _{xx}$ (1)

Suppose a rightward propagating wave exists within this waveguide,

$\psi (x,z)=\phi (z)\exp {\left(i(kx-\omega t)\right)}$,

then equation 1 becomes

$\phi _{zz}+\left({\frac {N^{2}k^{2}}{\omega ^{2}}}-k^{2}\right)\phi =0$. (2)

This is an eigenvalue problem which only has a simple solution when the stratification is linear (ie. $N^{2}(z)=N_{0}^{2}$). Equation (2) is then simplified to

$\phi _{zz}+m^{2}\phi =0$.

The solutions that satisfies both boundary conditions is

$\phi (z)=\sin(mz)$

where $m=n\pi /H$. The vertical wavenumber, $m$, is now quantized and only wave of a particular vertical structure exist. The vertical structure function, $\phi$, phase speed, $c_{p}$, and group speed, $c_{g}$ for the first three modes are plotted below