# Do I understand pressure coordinates in the right way?

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

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?

• 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. Mar 14, 2023 at 21:07
• 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). Mar 14, 2023 at 21:20
• 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. Mar 14, 2023 at 21:49
• @AtmosphericPrisonEscape the velocity naturally appears within the material derivative Mar 15, 2023 at 16:50
• 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. Mar 16, 2023 at 15:53

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:

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

$$$$\nabla_z A = \nabla_p A + \frac{\partial z}{\partial p} \frac{\partial A}{\partial z}\nabla_z p.$$$$ The time derivative is trivial as it follows from the general form above. Now we can write the material derivative in pressure coordinates as follows: $$$$\begin{split} \left( \frac{\text{D}}{\text{D} t} \right)_p &= \left( \frac{\partial}{\partial t} \right)_p+ \vec{u} \cdot \nabla_p + \omega \frac{\partial}{\partial p}\\ &= \left( \frac{\partial }{\partial t} \right)_z - \frac{\partial z}{\partial p}\left( \frac{\partial p}{\partial t} \right)_z \frac{\partial }{\partial z} + \vec{u} \cdot \left[ \nabla_z - \frac{\partial z}{\partial p} \nabla_z p \frac{\partial }{\partial z} \right] + \omega \frac{\partial z}{\partial p}\frac{\partial}{\partial z} \\ &= \left( \frac{\partial }{\partial t} \right)_z + \vec{u} \cdot \nabla_z + \left[\omega - \left(\frac{\partial p}{\partial t}\right)_z - \vec{u} \cdot \nabla_z p \right] \frac{\partial z}{\partial p} \frac{\partial}{\partial z} . \end{split}$$$$ Hence, $$$$w = \left[\omega - \left(\frac{\partial p}{\partial t}\right)_z - \vec{u} \cdot \nabla_z p \right] \frac{\partial z}{\partial p}$$$$ and $$$$\omega =\left( \frac{\partial p }{\partial t}\right)_z + \vec{u} \cdot \nabla_z p + w \frac{\partial p}{\partial z}.$$$$ What typically follows is the use of the hydrostatic approximation in the last term. If you do a (large) scale analysis you will find that $$-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$$.