<p>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 <a href="https://en.wikipedia.org/wiki/Stable_and_unstable_stratification" rel="nofollow noreferrer">stably stratified</a> 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.</p>
<p>The equations of motion for a spring are given by <span class="math-container">$$ m \frac{\partial^2 z}{\partial t^2} = -kz$$</span>, where <span class="math-container">$m$</span> is the mass of the spring, <span class="math-container">$z$</span> is the position relative to the resting position and <span class="math-container">$k$</span> is the spring constant. One can find the period of oscillation to be
<span class="math-container">$$T = {2 \pi} \sqrt{\frac{m}{k}}.$$</span></p>
<p>Now consider a parcel of air with density <span class="math-container">$\rho_0$</span> that is displaced (by an orographic lift), and that the surrounding air has a density <span class="math-container">$\rho(z)$</span>. We assume the density of our parcel to be constant. The force acting on that parcel is <span class="math-container">$F = -g(\rho_0-\rho(z))$</span> or
<span class="math-container">$${F} = \rho_0 \frac{\partial^2 z'}{\partial t^2} = -({\rho_0-\rho})g.$$</span> Notice that <span class="math-container">$z'$</span> is the displacement of the parcel (in contrast to the spring example above ) and <span class="math-container">$z$</span> is the vertical coordinate. We observe that the equations of motion for the spring and the air parcel are looking equivalent. However, since <span class="math-container">$\rho$</span> depends on <span class="math-container">$z$</span> we can't solve the equation without any further assumptions. If we assume <span class="math-container">$\rho(z) = \rho_0 + \frac{\partial \rho}{\partial z}z'$</span> we have
<span class="math-container">$$\rho_0 \frac{\partial^2 z'}{\partial t^2} = g\frac{\partial \rho}{\partial z}z'.$$</span> Once again comparing to the spring example we see that in this case the spring constant is <span class="math-container">$-g\frac{\partial \rho}{\partial z}$</span> and hence, the oscillation period is</p>
<p><span class="math-container">$$T = 2 \pi \sqrt{-\frac{g}{\rho_0}\frac{ \partial \rho}{ \partial z}}$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2024-06-20T13:06:14.843Z
2024-06-20T13:06:14.843Z
Joscha Fregin
<p>As the air descends the mountain, the gravitational potential energy is converted into thermal energy, causing it to heat up; As the air is warmer, it is less dense than the air in front of it, causing it to rush up over it.
The thermal energy converts to gravitational potential energy and the air flows down again</p>

CC BY-SA 4.0
2024-07-23T11:55:33.237Z
2024-07-23T11:55:33.237Z
Harrychink
<p>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?</p>

<atmosphere><wind><clouds><waves><orography>
How do lee waves form?
jack_O'Dim
# How do lee waves form?

(Asked by: username_0 on 2024-06-20)

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?

**Tags:** atmosphere, wind, clouds, waves, orography


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## 2 answers

### Answer by username_1 on 2024-06-20. Score: 2

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](https://en.wikipedia.org/wiki/Stable_and_unstable_stratification) 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.


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### Answer by username_2 on 2024-07-23. Score: 0

As the air descends the mountain, the gravitational potential energy is converted into thermal energy, causing it to heat up; As the air is warmer, it is less dense than the air in front of it, causing it to rush up over it.
The thermal energy converts to gravitational potential energy and the air flows down again


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