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Could anyone help me work through the following set of equations:

enter image description here

I found these online at

and I'm trying to work through them. I get the basics from this i.e. that if a water particle moves vertically it will be subjected to a buoyancy force due to it being surrounded by water of different density. However, I don't understand how the solution has worked i.e. how they went from one stage to the next. Could anyone provide more information about how this solution works?

My main aim here is to figure out how the Brunt vaisala buoyancy frequency is derived.

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Which part do you not understand? The force balance? Or the reduction to the simple harmonic oscillator? – Bernhard Feb 18 '13 at 16:17
I'm just struggling with making the links, such as when they state apply newtons law to the particle etc... – Emma Feb 18 '13 at 17:49
up vote 2 down vote accepted

Let start with Newton's second law in one dimension, then we get

$$\sum F_z = m \frac{\partial^2\zeta}{\partial t^2} $$

There are two forces acting on the particle

  • a downward force due to gravity: $m g = \rho(z) V g$
  • an upward buoyant force: $m_{displaced} g = \rho(z+\zeta) g$

Fill things in to get: $$\rho(z) \frac{\partial^2\zeta}{\partial t^2} = -(\rho(z)-\rho(z+\zeta) ) g $$

You can approximate the last term with a linearisation using a Taylor expansion


You see that the term containing $\rho(z)$ will cancel out, thus

$$\rho(z) \frac{\partial^2\zeta}{\partial t^2} = \frac{d\rho}{dz}\zeta g $$

Or, two rewrite it in the simple harmonic oscillator language

$$ \frac{\partial^2\zeta}{\partial t^2} - \left(\frac{g}{\rho(z)} \frac{d\rho}{dz}\right)\zeta = 0 $$

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