# How do lee waves form?

I know that they are caused (usually) by winds flowing up an orographic lift, but why does the up-and-down motion continue after surpassing the mountain?

## 1 Answer

It is similar to the effect of a spring. If you have mass attached to a spring and pull or push, such that the mass leaves its resting position, the spring will oscillate around the resting position. For lee waves to occur we need a stably stratified atmosphere. Because, otherwise there would, indeed, be no continuing up-and-down motion. The density difference between the fluid parcel that oscillates and the surrounding air then acts as the spring does in the introductory example. We can back up this qualitative explanation by very simple physics.

The equations of motion for a spring are given by $$m \frac{\partial^2 z}{\partial t^2} = -kz$$, where $$m$$ is the mass of the spring, $$z$$ is the position relative to the resting position and $$k$$ is the spring constant. One can find the period of oscillation to be $$T = {2 \pi} \sqrt{\frac{m}{k}}.$$

Now consider a parcel of air with density $$\rho_0$$ that is displaced (by an orographic lift), and that the surrounding air has a density $$\rho(z)$$. We assume the density of our parcel to be constant. The force acting on that parcel is $$F = -g(\rho_0-\rho(z))$$ or $${F} = \rho_0 \frac{\partial^2 z'}{\partial t^2} = -({\rho_0-\rho})g.$$ Notice that $$z'$$ is the displacement of the parcel (in contrast to the spring example above ) and $$z$$ is the vertical coordinate. We observe that the equations of motion for the spring and the air parcel are looking equivalent. However, since $$\rho$$ depends on $$z$$ we can't solve the equation without any further assumptions. If we assume $$\rho(z) = \rho_0 + \frac{\partial \rho}{\partial z}z'$$ we have $$\rho_0 \frac{\partial^2 z'}{\partial t^2} = g\frac{\partial \rho}{\partial z}z'.$$ Once again comparing to the spring example we see that in this case the spring constant is $$-g\frac{\partial \rho}{\partial z}$$ and hence, the oscillation period is

$$T = 2 \pi \sqrt{-\frac{g}{\rho_0}\frac{ \partial \rho}{ \partial z}}$$

What's left is to consider the cases where there is no oscillation which happens in a neutral or unstable stratification. If the atmosphere is neutral then there is no density difference and, hence, no force to push back the parcel. If the atmosphere is unstable the spring constant is negative, which means the spring keeps extending or shrinking. The air parcel will continue to rise.