<p>This is not a valid diagnostic equation.  The vertical velocity <span class="math-container">$\omega$</span> is defined in vertical pressure <span class="math-container">$p$</span> coordinate as <span class="math-container">$\omega = Dp/Dt$</span>.  As a result, all the equations you used should be in pressure coordinate.</p>
<p>Then the continuity equation
<span class="math-container">$$D\rho/Dt=-\rho\nabla\cdot\vec u$$</span></p>
<p>will change to
<span class="math-container">$$\nabla_p\cdot\vec u=\partial_x u+\partial_y v+\partial_p \omega=0$$</span></p>
<p>Also, density <span class="math-container">$\rho$</span> in pressure coordinate becomes a constant <span class="math-container">$1/g$</span>.  You should use everything in pressure coordinate for a diagnostic equation of <span class="math-container">$\omega$</span>.</p>

CC BY-SA 4.0
2022-11-12T16:10:45.577Z
2022-11-13T13:19:48.457Z
2022-11-13T13:19:48.457Z
MiniUFO
<p>Beginning with the Primitive equations governing atmospheric motion for a dry gas, primarily the ideal gas law, and the conservation of mass and energy, neglecting diffusivity.</p>

<p>$$P=\rho R T$$ 
$$\omega \equiv \frac{DP}{Dt}$$
Therefore
$$\omega=R(\rho\frac{DT}{Dt}+T\frac{D\rho}{Dt})$$
Since $$\frac{D\rho}{Dt}=-\rho \nabla\cdot\vec{u}$$ and $$\frac{DT}{Dt}=\frac{\omega\rho}{c_p}+\frac{Q}{c_p}$$
$$\omega=R(\frac{\rho^2\omega}{c_p}+\frac{\rho Q}{c_p}-T\rho\nabla\cdot\vec{u})$$
Distributing $R$ and applying the ideal gas law $$\omega=\frac{R\rho^2\omega}{c_p}+\frac{PQ}{Tc_p}-P\nabla\cdot\vec{u}$$
Separating $\omega$ from the right hand side of the equation yields
$$\omega=(1-\frac{R\rho^2}{c_p})^{-1}(\frac{PQ}{Tc_p}-P\nabla\cdot\vec{u})$$</p>

<p>I call this semi-diagnostic, for $\omega$ is still a part of $\nabla\cdot\vec{u}$</p>

<p>Is this a valid semi-diagnostic equation for $\omega$?</p>

CC BY-SA 3.0
2017-03-31T15:26:32.370Z
Being that I didn't use the Navier-Stokes equations, there is no need to assume a momentum balance. If no, what did I do wrong? I am sure there is an alternate derivation- the final equation can be interpreted as a restatement of the conservation of energy with diabatic heating and work on the right hand side.
BarocliniCplusplus
CC BY-SA 3.0
2017-04-01T18:00:46.593Z
Here is the chat room: http://chat.stackexchange.com/rooms/56422/semi-diagnostic-omega
BarocliniCplusplus
CC BY-SA 3.0
2017-04-07T17:34:14.633Z
To clarify, I am using the Conservation of energy, the conservation of mass, and the ideal gas law (a subset of the primitive equations) to derive omega.
BarocliniCplusplus
CC BY-SA 3.0
2017-03-09T21:08:12.283Z
Been a long while since I've worked in derivations and such, and so I may not be the best source of input (it gets pretty complex for me as such towards the end!).  But saying rho changes over time only due to advection seems a significant simplification?
JeopardyTempest
CC BY-SA 3.0
2017-03-09T21:09:18.000Z
(you may want to give more of the assumptions\sources of some of those equations around that point)
JeopardyTempest
CC BY-SA 3.0
2017-03-09T21:36:20.147Z
Sure, the continuity assumption and the ideal gas assumption, are two base assumptions. I also assume a dry gas. The capital D/Dt denotes the Lagrangian derivative, such that advection is built into the derivative. The continuity equation in nonconservative form, and my definition for conservation of energy (loosely) is from "An Introduction to Dynamic Meteorology" by Holton. My definition for omega comes from "Atmospheric Science" by Wallace and Hobbes.
BarocliniCplusplus
<thermodynamics><dynamics>
Is this a valid semi-diagnostic equation for Omega?
BarocliniCplusplus
# Is this a valid semi-diagnostic equation for Omega?

(Asked by: username_0 on 2017-03-09)

Beginning with the Primitive equations governing atmospheric motion for a dry gas, primarily the ideal gas law, and the conservation of mass and energy, neglecting diffusivity.

$$P=\rho R T$$
$$\omega \equiv \frac{DP}{Dt}$$
Therefore
$$\omega=R(\rho\frac{DT}{Dt}+T\frac{D\rho}{Dt})$$
Since $$\frac{D\rho}{Dt}=-\rho \nabla\cdot\vec{u}$$ and $$\frac{DT}{Dt}=\frac{\omega\rho}{c\_p}+\frac{Q}{c\_p}$$
$$\omega=R(\frac{\rho^2\omega}{c\_p}+\frac{\rho Q}{c\_p}-T\rho\nabla\cdot\vec{u})$$
Distributing $R$ and applying the ideal gas law $$\omega=\frac{R\rho^2\omega}{c\_p}+\frac{PQ}{Tc\_p}-P\nabla\cdot\vec{u}$$
Separating $\omega$ from the right hand side of the equation yields
$$\omega=(1-\frac{R\rho^2}{c\_p})^{-1}(\frac{PQ}{Tc\_p}-P\nabla\cdot\vec{u})$$

I call this semi-diagnostic, for $\omega$ is still a part of $\nabla\cdot\vec{u}$

Is this a valid semi-diagnostic equation for $\omega$?

**Tags:** thermodynamics, dynamics

> **Comments:**
> **username_1**: Been a long while since I've worked in derivations and such, and so I may not be the best source of input (it gets pretty complex for me as such towards the end!).  But saying rho changes over time only due to advection seems a significant simplification?  
> **username_1**: (you may want to give more of the assumptions\sources of some of those equations around that point)  
> **username_0**: Sure, the continuity assumption and the ideal gas assumption, are two base assumptions. I also assume a dry gas. The capital D/Dt denotes the Lagrangian derivative, such that advection is built into the derivative. The continuity equation in nonconservative form, and my definition for conservation of energy (loosely) is from "An Introduction to Dynamic Meteorology" by Holton. My definition for omega comes from "Atmospheric Science" by Wallace and Hobbes.  
> **username_0**: Being that I didn't use the Navier-Stokes equations, there is no need to assume a momentum balance. If no, what did I do wrong? I am sure there is an alternate derivation- the final equation can be interpreted as a restatement of the conservation of energy with diabatic heating and work on the right hand side.  
> **username_0**: Here is the chat room: http://chat.stackexchange.com/rooms/56422/semi-diagnostic-omega  
> **username_0**: To clarify, I am using the Conservation of energy, the conservation of mass, and the ideal gas law (a subset of the primitive equations) to derive omega.  

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## 1 answers

### Answer by username_2 on 2022-11-12. Score: 0 | Accepted answer

This is not a valid diagnostic equation. The vertical velocity $\omega$ is defined in vertical pressure $p$ coordinate as $\omega = Dp/Dt$. As a result, all the equations you used should be in pressure coordinate.

Then the continuity equation
$$D\rho/Dt=-\rho\nabla\cdot\vec u$$

will change to
$$\nabla\_p\cdot\vec u=\partial\_x u+\partial\_y v+\partial\_p \omega=0$$

Also, density $\rho$ in pressure coordinate becomes a constant $1/g$. You should use everything in pressure coordinate for a diagnostic equation of $\omega$.


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