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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 $\omega$?

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  • $\begingroup$ 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? $\endgroup$ Mar 9, 2017 at 21:08
  • $\begingroup$ (you may want to give more of the assumptions\sources of some of those equations around that point) $\endgroup$ Mar 9, 2017 at 21:09
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    $\begingroup$ 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. $\endgroup$ Mar 9, 2017 at 21:36
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    $\begingroup$ 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. $\endgroup$ Mar 31, 2017 at 15:26
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    $\begingroup$ Here is the chat room: chat.stackexchange.com/rooms/56422/semi-diagnostic-omega $\endgroup$ Apr 1, 2017 at 18:00

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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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