<p>There is a misconception in your assumption:</p>
<p><span class="math-container">$\omega = \frac{Dp}{Dt} = 0 $</span> means that the mass above the air parcel stays constant.</p>
<blockquote>
<p>Proof: Assume quasistatic eqn:</p>
<p>-<span class="math-container">$\frac{1}{g}\, dp=\rho \, dz$</span>;</p>
<p>apply operator of horizontal mean: <span class="math-container">$\overline{(\cdot)}= \frac{\int (\cdot)  \, dxdy}{\int dxdy}=\frac{\int (\cdot)  \, dA}{\Delta A}$</span></p>
<p>-<span class="math-container">$\frac{1}{g}\, \overline{dp} \, \Delta A=\rho \, dxdydz= \rho dV = dM$</span></p>
<p>Integrating between arbitrary pressure levels gives mass between those levels.</p>
</blockquote>
<p>(1) You correctly estimated <span class="math-container">$\partial p_s/\partial y &lt;0 $</span> in order to have wind to the north, but</p>
<p>(2) you also set <span class="math-container">$\partial p_s/\partial t =0 $</span> in your derivation (to replace total derivative with advection of <span class="math-container">$p_s$</span>). This means that the pressure field is constant in time.</p>
<h1>Result</h1>
<p>(1) and (2) yield to lower pressure in the north (<span class="math-container">$dp_s/dy&lt;0$</span>) and decreasing surface pressure following the parcel (<span class="math-container">$dp_s/dt&lt;0$</span>)</p>
<p>This means that an air parcel has to descend (in terms of <span class="math-container">$\sigma$</span>) in order to satisfy the condition of constant mass lying over it (<span class="math-container">$\omega = \frac{Dp}{Dt} = 0 $</span>):</p>
<p><span class="math-container">$d\sigma/dt = a \, dp/dt- b\, dp_s/dt$</span>; (with a,b const.)</p>
<p>=&gt; <span class="math-container">$d\sigma/dt &gt;0$</span></p>
<h1>Example</h1>
<p>If your parcel starts at 950 hPa, travels to north with falling surface pressure, condition <span class="math-container">$\omega=0$</span> means it stays at 950 hPa, but with (1) surface pressure drops to 950 hPa as the parcel continues north and the parcel miraculously is found to be at ground level.</p>
<h1>Conclusion</h1>
<p>As <span class="math-container">$\sigma$</span> depends on pressure and surface pressure, so does its time derivative.</p>

CC BY-SA 3.0
2015-12-29T03:29:01.117Z
2015-12-29T03:29:01.117Z
2020-06-18T08:25:19.090Z
Lukas
<p>I have a problem in deriving/understanding the vertical velocity in sigma-pressure coordinates.</p>

<p>The vertical coordinate is defined as:</p>

<p>$$\sigma(x,y,p)=\frac{p(x,y)-p_t}{p_s(x,y)-p_t}$$</p>

<p>where $p(x,y)$ is the pressure of the considered grid point, $p_t$ is the pressure at the top of the domain (in my case $50$hPa) and $p_s(x,y)$ is the surface pressure.</p>

<p>After some derivation I get for the vertical velocity ($D/Dt$ is the total, $\partial/\partial t$ the partial derivative, $\omega = Dp/Dt$):</p>

<p>$$w = \frac{D \sigma}{Dt} = - \frac{\sigma}{(p_s-p_t)}  \left(\frac{\partial p_s}{\partial x} u + \frac{\partial p_s}{\partial y} v \right) + \frac{1}{p_s-p_t} \omega$$</p>

<p>When I now think of an idealized example where I have a gaussian hill in the middle of the domain and wind coming only from the south ($v=10$ms$^{-1}$, $u$, $\omega=0$), the resulting vertical wind component is positive in front of the hill which is wrong since the air should be ascending (positive value of $w$ means descending).</p>

<p>Where is my mistake?</p>

CC BY-SA 3.0
2015-10-09T00:01:17.303Z
Is there time dependence in your equation for $\sigma$ ? And what is $\omega$ ? It is not clear to me how it is defined as the full derivative of $p$ with respect to $t$. Isn't $p$ defined as $p(x,y)$?
Isopycnal Oscillation
CC BY-SA 3.0
2015-10-09T17:18:55.870Z
@gansub ok its the material derivative, makes more sense. I'll edit it.
Isopycnal Oscillation
<meteorology>
How is the vertical velocity defined in sigma-pressure-coordinates?
ConfusedMeteorologist
# How is the vertical velocity defined in sigma-pressure-coordinates?

(Asked by: username_0 on 2015-10-07)

I have a problem in deriving/understanding the vertical velocity in sigma-pressure coordinates.

The vertical coordinate is defined as:

$$\sigma(x,y,p)=\frac{p(x,y)-p\_t}{p\_s(x,y)-p\_t}$$

where $p(x,y)$ is the pressure of the considered grid point, $p\_t$ is the pressure at the top of the domain (in my case $50$hPa) and $p\_s(x,y)$ is the surface pressure.

After some derivation I get for the vertical velocity ($D/Dt$ is the total, $\partial/\partial t$ the partial derivative, $\omega = Dp/Dt$):

$$w = \frac{D \sigma}{Dt} = - \frac{\sigma}{(p\_s-p\_t)} \left(\frac{\partial p\_s}{\partial x} u + \frac{\partial p\_s}{\partial y} v \right) + \frac{1}{p\_s-p\_t} \omega$$

When I now think of an idealized example where I have a gaussian hill in the middle of the domain and wind coming only from the south ($v=10$ms$^{-1}$, $u$, $\omega=0$), the resulting vertical wind component is positive in front of the hill which is wrong since the air should be ascending (positive value of $w$ means descending).

Where is my mistake?

**Tags:** meteorology

> **Comments:**
> **username_1**: Is there time dependence in your equation for $\sigma$ ? And what is $\omega$ ? It is not clear to me how it is defined as the full derivative of $p$ with respect to $t$. Isn't $p$ defined as $p(x,y)$?  
> **username_1**: @gansub ok its the material derivative, makes more sense. I'll edit it.  

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## 1 answers

### Answer by username_2 on 2015-12-29. Score: 3

There is a misconception in your assumption:

$\omega = \frac{Dp}{Dt} = 0 $ means that the mass above the air parcel stays constant.

> Proof: Assume quasistatic eqn:
>
> -$\frac{1}{g}\, dp=\rho \, dz$;
>
> apply operator of horizontal mean: $\overline{(\cdot)}= \frac{\int (\cdot) \, dxdy}{\int dxdy}=\frac{\int (\cdot) \, dA}{\Delta A}$
>
> -$\frac{1}{g}\, \overline{dp} \, \Delta A=\rho \, dxdydz= \rho dV = dM$
>
> Integrating between arbitrary pressure levels gives mass between those levels.

(1) You correctly estimated $\partial p\_s/\partial y <0 $ in order to have wind to the north, but

(2) you also set $\partial p\_s/\partial t =0 $ in your derivation (to replace total derivative with advection of $p\_s$). This means that the pressure field is constant in time.

# Result

(1) and (2) yield to lower pressure in the north ($dp\_s/dy<0$) and decreasing surface pressure following the parcel ($dp\_s/dt<0$)

This means that an air parcel has to descend (in terms of $\sigma$) in order to satisfy the condition of constant mass lying over it ($\omega = \frac{Dp}{Dt} = 0 $):

$d\sigma/dt = a \, dp/dt- b\, dp\_s/dt$; (with a,b const.)

=> $d\sigma/dt >0$

# Example

If your parcel starts at 950 hPa, travels to north with falling surface pressure, condition $\omega=0$ means it stays at 950 hPa, but with (1) surface pressure drops to 950 hPa as the parcel continues north and the parcel miraculously is found to be at ground level.

# Conclusion

As $\sigma$ depends on pressure and surface pressure, so does its time derivative.


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