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.


1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.