Internal Wave: Difference between revisions

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(Created page with "== 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 <math> ...")
 
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:<math> (\nabla^2 \psi)_{tt} = - N^2(z) \psi_{xx} </math>.
:<math> (\nabla^2 \psi)_{tt} = - N^2(z) \psi_{xx} </math>.
Making the wave ansatz, <math> \psi = \exp{\left(\vec{k}\cdot\vec{x} - \omega t\right)} </math>, where <math> \vec{k}=(k_x,k_z)</math> and <math> \vec{x}=(x,z)</math>, the  dispersion relation is
Making the wave ansatz, <math> \psi = \exp{\left(\vec{k}\cdot\vec{x} - \omega t\right)} </math>, where <math> \vec{k}=(k_x,k_z)</math> and <math> \vec{x}=(x,z)</math>, the  dispersion relation is
:<math> \omega = \frac{N k_x}{\lvert\vec{k}\rvert} </math>.
:<math> \omega = \frac{N k_x}{\lvert\vec{k}\rvert} </math>
where <math> \lvert\vec{k}\rvert = \sqrt{k_x^2 + k_z^2} </math>.




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Therefore we find that the group velocity is perpendicular to the phase velocity!
Therefore we find that the group velocity is perpendicular to the phase velocity!
:<math> 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</math>
:<math> 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</math>
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 ==
== Waves in a channel ==

Revision as of 11:02, 23 June 2015

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 coordinate. The Navier-Stokes equations can therefore be written using the stream-function, , as

.

Making the wave ansatz, , where and , the dispersion relation is

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

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lvert\vec{k}\rvert = \sqrt{k_x^2 + k_z^2} } .


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 (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}

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