<ol>
<li><p>In pressure coordinates it is common to use <span class="math-container">$\omega$</span> as the vertical velocity component instead of <span class="math-container">$w$</span> as in cartesian coordinates.</p>
</li>
<li><p>Obtaining the velocity components in the pressure coordinate system:</p>
</li>
</ol>
<p>Using the material derivative we can relate the vertical velocity components and I will describe below how to do that.</p>
<p>Start with two coordinate systems <span class="math-container">$(x, y, z, t)$</span> and <span class="math-container">$(x, y, p, t)$</span> with vertical coordinates <span class="math-container">$z = z(t, x, y, p)$</span> and <span class="math-container">$p = p(t, x, y, z)$</span>. A derivative of some variable <span class="math-container">$A$</span> with respect to some other variable <span class="math-container">$c$</span> (which can be <span class="math-container">$x$</span>, <span class="math-container">$y$</span> or <span class="math-container">$t$</span>) in one system is related to the other as follows:</p>
<p><span class="math-container">\begin{equation}
\left( \frac{\partial A}{\partial c} \right)_z = \left( \frac{\partial A}{\partial c} \right)_p + \frac{\partial A}{\partial p}\left( \frac{\partial p}{\partial c} \right)_z.
\end{equation}</span>
For the vertical coordinates it holds that
<span class="math-container">\begin{equation}
\frac{\partial A}{\partial p} =  \frac{\partial A}{\partial z} \frac{\partial z}{\partial p},
\end{equation}</span>
and thus, we find
<span class="math-container">\begin{equation}
\left( \frac{\partial A}{\partial c} \right)_z = \left( \frac{\partial A}{\partial c} \right)_p +  \frac{\partial A}{\partial z} \frac{\partial z}{\partial p}\left( \frac{\partial p}{\partial c} \right)_z.
\end{equation}</span></p>
<p>With this the (2D) gradient reads:</p>
<p><span class="math-container">\begin{equation}
 \nabla_z A = \nabla_p A + \frac{\partial z}{\partial p} \frac{\partial A}{\partial z}\nabla_z p.
\end{equation}</span>
The time derivative is trivial as it follows from the general form above. Now we can write the material derivative in pressure coordinates as follows:
<span class="math-container">\begin{equation}
\begin{split}
\left( \frac{\text{D}}{\text{D} t} \right)_p &amp;= \left( \frac{\partial}{\partial t} \right)_p+ \vec{u} \cdot \nabla_p + \omega \frac{\partial}{\partial p}\\
&amp;= \left( \frac{\partial }{\partial t} \right)_z -  \frac{\partial z}{\partial p}\left( \frac{\partial p}{\partial t} \right)_z \frac{\partial }{\partial z} + \vec{u} \cdot \left[ \nabla_z  - \frac{\partial z}{\partial p} \nabla_z p \frac{\partial }{\partial z} \right] + \omega \frac{\partial z}{\partial p}\frac{\partial}{\partial z} \\
&amp;= \left( \frac{\partial }{\partial t} \right)_z + \vec{u} \cdot \nabla_z + \left[\omega - \left(\frac{\partial p}{\partial t}\right)_z - \vec{u} \cdot \nabla_z p \right] \frac{\partial z}{\partial p} \frac{\partial}{\partial z} .
\end{split}
\end{equation}</span>
Hence,
<span class="math-container">\begin{equation}
 w = \left[\omega - \left(\frac{\partial p}{\partial t}\right)_z - \vec{u} \cdot \nabla_z p \right] \frac{\partial z}{\partial p}
\end{equation}</span>
and
<span class="math-container">\begin{equation}
\omega =\left( \frac{\partial p }{\partial t}\right)_z + \vec{u} \cdot \nabla_z p + w \frac{\partial p}{\partial z}.
\end{equation}</span>
What typically follows is the use of the hydrostatic approximation in the last term. If you do a (large) scale analysis you will find that <span class="math-container">$-w\rho g = w \partial p/\partial z$</span> is the dominating term and thus, you can approximate <span class="math-container">$\omega = -w\rho g$</span>. However, based on your sketch I think it would not add to an understanding to just say <span class="math-container">$\omega = 0$</span>, since <span class="math-container">$w = 0$</span>.</p>
<p>Why it makes sense to use pressure coordinates I tried to answer <a href="https://earthscience.stackexchange.com/questions/19754/what-is-the-explanation-of-using-pressure-units-hpa-to-express-height-when-dea">here</a>, although I want to add that in a theoretical context it can be much more useful to apply e.g. sigma coordinates (terrain following coordinates) to simplify the treatment of boundary conditions.</p>

CC BY-SA 4.0
2023-03-15T16:40:50.250Z
2023-03-16T17:12:33.820Z
2023-03-16T17:12:33.820Z
Joscha Fregin
<p>The question is simple: I still have problems with atmospheric pressure coordinates:</p>
<p><a href="https://i.sstatic.net/FDpSF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FDpSF.png" alt="enter image description here" /></a></p>
<p>Given the pressure dependence in the picture, what are the velocity coordinates of an air parcel moving with <span class="math-container">$\vec v =(u, v=0, w=0)$</span> along zonal direction in <strong>pressure coordinates</strong>?</p>
<p>Do I understand right, that in pressure coordinates there is no w but ω instead?</p>
<p>I would say we have then the tuple
<span class="math-container">$$(u, v=0, ω)$$</span>
with
<span class="math-container">$$\omega = Dp/Dt = (\partial p/\partial x)_z \cdot Dx/Dt= (\partial p/\partial x)_z \cdot u$$</span></p>
<p>But I'm not sure, because it seems a bit weird that in the first system we have zero w and in the pressure system we have negative vertical &quot;speed&quot;.</p>
<p>Although we normally consider only horizontal components u, v of wind, due to pressure coordinates we have a non zero &quot;vertical&quot; component (because pressure decreases as the parcel moves to the right in the example). How are pressure coordinates than easier to handle when there is an additional component <span class="math-container">$\omega$</span> and we have three velocity components to consider instead of just two (of course I know that other equations become more simple as density vanishes...)?</p>
<p>Do I have a mistake in my considerations?</p>

CC BY-SA 4.0
2023-03-14T21:07:26.117Z
I don't think it is correct to invoke any velocity in the pressure coordinate transformation, as the pressure coordinate system must also work in a hydrostatic setting. I always understood $p(x,z) = const.$ as encoding some unknown function $f(x,z) \propto p(x,z)$. I can read off $f(x,z)$ in this case from the graph, and use its differentials w.r.t x and z in order to integrate the actual function for $p(x,z)$. When you know the relation $p(x,z)$, all is said and done, because the differential $\partial p/\partial z$ is the transformation law that you are looking for.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-03-14T21:20:01.687Z
But in my example the air parcel experiences decreasing pressure as it moves from left to right in constant geopotential height. So its pressure coordinate changes - right? What is $\omega$ then? In the lecture we learned $\omega = Dp/Dt$ and in this case it would be negative (assuming hydrostatic approximation).
MichaelW
CC BY-SA 4.0
2023-03-14T21:49:38.143Z
That's just the definition of a velocity, not a coordinate transformation.... you need to know the form of your transformation in order to transform your fundamental equations (such as Euler/Navier stokes). After this is done, you can do all other shenanigans, but in pressure coordinates.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-03-15T16:50:11.210Z
@AtmosphericPrisonEscape the velocity naturally appears within the material derivative
Joscha Fregin
CC BY-SA 4.0
2023-03-15T21:23:06.187Z
@JoschaFregin Just that the material derivative has nothing to do with a proper coordinate transform. In your answer, you are obscuring what is happening. As I said, pressure coordinates have to work in a hydrostatic setting, which you conveniently ignored.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-03-16T15:53:37.540Z
The more I think the more I realise that my answer is pretty stupid. I removed the example as it is probably more confusing than helpful but added a few lines on how one can relate $w$ and $\omega$ via the material derivative - as it appears in the question. My answer still needs quite some work but maybe it is at least a little helpful. I guess now would be a good time for me to get back at reading.
Joscha Fregin
<meteorology><atmosphere><atmosphere-modelling><pressure><coordinate-system>
Do I understand pressure coordinates in the right way?
MichaelW
# Do I understand pressure coordinates in the right way?

(Asked by: username_0 on 2023-03-14)

The question is simple: I still have problems with atmospheric pressure coordinates:

[![enter image description here](https://i.sstatic.net/FDpSF.png)](https://i.sstatic.net/FDpSF.png)

Given the pressure dependence in the picture, what are the velocity coordinates of an air parcel moving with $\vec v =(u, v=0, w=0)$ along zonal direction in **pressure coordinates**?

Do I understand right, that in pressure coordinates there is no w but ω instead?

I would say we have then the tuple
$$(u, v=0, ω)$$
with
$$\omega = Dp/Dt = (\partial p/\partial x)\_z \cdot Dx/Dt= (\partial p/\partial x)\_z \cdot u$$

But I'm not sure, because it seems a bit weird that in the first system we have zero w and in the pressure system we have negative vertical "speed".

Although we normally consider only horizontal components u, v of wind, due to pressure coordinates we have a non zero "vertical" component (because pressure decreases as the parcel moves to the right in the example). How are pressure coordinates than easier to handle when there is an additional component $\omega$ and we have three velocity components to consider instead of just two (of course I know that other equations become more simple as density vanishes...)?

Do I have a mistake in my considerations?

**Tags:** meteorology, atmosphere, atmosphere-modelling, pressure, coordinate-system

> **Comments:**
> **username_1**: I don't think it is correct to invoke any velocity in the pressure coordinate transformation, as the pressure coordinate system must also work in a hydrostatic setting. I always understood $p(x,z) = const.$ as encoding some unknown function $f(x,z) \propto p(x,z)$. I can read off $f(x,z)$ in this case from the graph, and use its differentials w.r.t x and z in order to integrate the actual function for $p(x,z)$. When you know the relation $p(x,z)$, all is said and done, because the differential $\partial p/\partial z$ is the transformation law that you are looking for.  
> **username_0**: But in my example the air parcel experiences decreasing pressure as it moves from left to right in constant geopotential height. So its pressure coordinate changes - right? What is $\omega$ then? In the lecture we learned $\omega = Dp/Dt$ and in this case it would be negative (assuming hydrostatic approximation).  
> **username_1**: That's just the definition of a velocity, not a coordinate transformation.... you need to know the form of your transformation in order to transform your fundamental equations (such as Euler/Navier stokes). After this is done, you can do all other shenanigans, but in pressure coordinates.  
> **username_2**: @username_1 the velocity naturally appears within the material derivative  
> **username_1**: @username_2 Just that the material derivative has nothing to do with a proper coordinate transform. In your answer, you are obscuring what is happening. As I said, pressure coordinates have to work in a hydrostatic setting, which you conveniently ignored.  
> **username_2**: The more I think the more I realise that my answer is pretty stupid. I removed the example as it is probably more confusing than helpful but added a few lines on how one can relate $w$ and $\omega$ via the material derivative - as it appears in the question. My answer still needs quite some work but maybe it is at least a little helpful. I guess now would be a good time for me to get back at reading.  

---

## 1 answers

### Answer by username_2 on 2023-03-15. Score: 3 | Accepted answer

1. In pressure coordinates it is common to use $\omega$ as the vertical velocity component instead of $w$ as in cartesian coordinates.
2. Obtaining the velocity components in the pressure coordinate system:

Using the material derivative we can relate the vertical velocity components and I will describe below how to do that.

Start with two coordinate systems $(x, y, z, t)$ and $(x, y, p, t)$ with vertical coordinates $z = z(t, x, y, p)$ and $p = p(t, x, y, z)$. A derivative of some variable $A$ with respect to some other variable $c$ (which can be $x$, $y$ or $t$) in one system is related to the other as follows:

\begin{equation}
\left( \frac{\partial A}{\partial c} \right)\_z = \left( \frac{\partial A}{\partial c} \right)\_p + \frac{\partial A}{\partial p}\left( \frac{\partial p}{\partial c} \right)\_z.
\end{equation}
For the vertical coordinates it holds that
\begin{equation}
\frac{\partial A}{\partial p} = \frac{\partial A}{\partial z} \frac{\partial z}{\partial p},
\end{equation}
and thus, we find
\begin{equation}
\left( \frac{\partial A}{\partial c} \right)\_z = \left( \frac{\partial A}{\partial c} \right)\_p + \frac{\partial A}{\partial z} \frac{\partial z}{\partial p}\left( \frac{\partial p}{\partial c} \right)\_z.
\end{equation}

With this the (2D) gradient reads:

\begin{equation}
\nabla\_z A = \nabla\_p A + \frac{\partial z}{\partial p} \frac{\partial A}{\partial z}\nabla\_z p.
\end{equation}
The time derivative is trivial as it follows from the general form above. Now we can write the material derivative in pressure coordinates as follows:
\begin{equation}
\begin{split}
\left( \frac{\text{D}}{\text{D} t} \right)\_p &= \left( \frac{\partial}{\partial t} \right)\_p+ \vec{u} \cdot \nabla\_p + \omega \frac{\partial}{\partial p}\\
&= \left( \frac{\partial }{\partial t} \right)\_z - \frac{\partial z}{\partial p}\left( \frac{\partial p}{\partial t} \right)\_z \frac{\partial }{\partial z} + \vec{u} \cdot \left[ \nabla\_z - \frac{\partial z}{\partial p} \nabla\_z p \frac{\partial }{\partial z} \right] + \omega \frac{\partial z}{\partial p}\frac{\partial}{\partial z} \\
&= \left( \frac{\partial }{\partial t} \right)\_z + \vec{u} \cdot \nabla\_z + \left[\omega - \left(\frac{\partial p}{\partial t}\right)\_z - \vec{u} \cdot \nabla\_z p \right] \frac{\partial z}{\partial p} \frac{\partial}{\partial z} .
\end{split}
\end{equation}
Hence,
\begin{equation}
w = \left[\omega - \left(\frac{\partial p}{\partial t}\right)\_z - \vec{u} \cdot \nabla\_z p \right] \frac{\partial z}{\partial p}
\end{equation}
and
\begin{equation}
\omega =\left( \frac{\partial p }{\partial t}\right)\_z + \vec{u} \cdot \nabla\_z p + w \frac{\partial p}{\partial z}.
\end{equation}
What typically follows is the use of the hydrostatic approximation in the last term. If you do a (large) scale analysis you will find that $-w\rho g = w \partial p/\partial z$ is the dominating term and thus, you can approximate $\omega = -w\rho g$. However, based on your sketch I think it would not add to an understanding to just say $\omega = 0$, since $w = 0$.

Why it makes sense to use pressure coordinates I tried to answer [here](https://earthscience.stackexchange.com/questions/19754/what-is-the-explanation-of-using-pressure-units-hpa-to-express-height-when-dea), although I want to add that in a theoretical context it can be much more useful to apply e.g. sigma coordinates (terrain following coordinates) to simplify the treatment of boundary conditions.


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