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My questions relate to page 29 of this document. In particular, on page 29, two expressions for the bulk Richardson number are given; one is said to be 'before scaling'. How is the first expression for the bulk Richardson number $Ri_0 = gh/\rho_0$ derived from the unscaled $Ri_0 = g\Delta\rho h/(\rho_0 \Delta U^2)$? I can't seem to find a way of linking the two. Also, how is the phase speed of the internal gravity wave derived as $c = \sqrt{Ri_0/\alpha}$? I can't seem to figure out how this was obtained. Any help with the above questions is much appreciated.

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I am not sure but there might be a small error in this derivation that compensates itself. It has been quite some time since I have seen a linear stability analysis so take the following thoughts with a pinch of salt.

The Richardson number is defined as

$$ Ri := \frac{g}{\rho} \frac{\frac{\partial \rho}{\partial z}}{\left( \frac{\partial u}{\partial z} \right)^2} \phantom{spacespace} \frac{\text{buoyancy}}{\text{flow shear}}. \tag{1}\label{1}$$

When approximating the derivatives in \eqref{1} as finite differences (bulk Richardson number) this leads to

$$ Ri = \frac{g}{\rho} \frac{\frac{\Delta \rho}{\Delta z}}{\left( \frac{\Delta u}{\Delta z} \right)^2} = \frac{g}{\rho} \frac{\Delta \rho \Delta z}{\Delta u^2}. \tag{2}\label{2}$$

We transform the velocity to $\overline{U} = \frac{U_1 + U_2}{2}$ and for convenience introduce $\Delta z = h$ (see page 26) for the parameter for velocity transition between 1 and 2 which is assumed linear and of finite thinkness. Thus \eqref{2} can also be written for the point $z = 0$ (thus the index $0$) when considering the intervals I and II as

$$ Ri_0 = \frac{g}{\rho_0} \frac{\Delta \rho h}{\Delta \overline{U}^2}. \tag{3}\label{3} $$

Now we consider a scaled system where the all differences are normalised to the range $\left[ -1, 1 \right]$ such as in the figure on page 29 and therefore all the changes are $1 - (-1) = 2$ ($\Delta U = 2$ and $\Delta \rho = 2$). This allows us to simplify \eqref{3} to

$$ Ri_0 = \frac{g h}{2 \rho_0} \tag{4}\label{4} $$

which is referred to as "after scaling". Contrary to the document I still got a factor of two in there but it eliminates itself later on.

The phase speed is given as (page 5)

$$ c_p = \frac{\omega}{k} \tag{5}\label{5} $$

and the wave number by (page 5)

$$ \omega^2 = \frac{g ( \rho_2 - \rho_1) k}{\rho_1 + \rho_2} \tag{6}\label{6} $$

When transforming \eqref{6} to the scaled system $\rho_{1,2} = \rho_0 \pm 1$ this results in

$$ \omega^2 = \frac{g \overbrace{[\rho_0 + 1 - (\rho_0 - 1)]}^{2} k}{\underbrace{ \rho_0 + 1 + \rho_0 - 1}_{2 \rho_0} } = \frac{g k}{\rho_0} \tag{7}\label{7}$$

and therefore combining \eqref{6} and \eqref{7} $c_p$ is given by

$$ c_p = \frac{1}{k^2} \sqrt{\frac{g k}{\rho_0}} = \sqrt{\underbrace{\frac{g h}{2 \rho_0}}_{Ri_0} \underbrace{\frac{2}{k h}}_{\frac{1}{\alpha}}} = \sqrt{\frac{Ri_0}{\alpha}} \tag{8}\label{8} $$

as (page 26) $ \alpha = \frac{k h}{2}$ holds.

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  • $\begingroup$ Thank you very much for this explanation, it was very helpful. There does indeed appear to be some errors and ambiguity in the text, especially around scaling. Progressing through Chapter 2 of the same document, I came unstuck on a related calculation (page 41), where a dispersion relation is written in terms of a new constant 'R'. I'd be interested to see what you make of this, as (once again) I cannot seem to get their answer (the second equation on page 41). Any help with this is again much appreciated. $\endgroup$
    – wrb98
    Nov 5, 2019 at 20:34
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    $\begingroup$ I've added a link to the specific question: physics.stackexchange.com/questions/512233/… $\endgroup$
    – wrb98
    Nov 6, 2019 at 1:53
  • $\begingroup$ You are welcome! Glad it helped. I will have a look at it later today. Off to bed for now. $\endgroup$
    – 2b-t
    Nov 6, 2019 at 1:55
  • $\begingroup$ Thank you, is there any chance you could look at the other problem soon? $\endgroup$
    – wrb98
    Nov 10, 2019 at 17:06
  • $\begingroup$ I did so as promised the next day but I have no idea what happened in between those two lines either... I guess I would have to go through the entire document but I sadly don't have time for this at the moment. Sorry I could not help. $\endgroup$
    – 2b-t
    Nov 10, 2019 at 17:55

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