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In the book by Jonathan E. Martin "Mid-Latitude Atmospheric Dynamics - A First Course"

the continuity equation for isentropic coordinates is derived by applying conservation of mass:

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$\frac{d}{dt}(\frac{\partial p}{\partial \theta}) + \frac{\partial p}{\partial \theta} \frac{\partial u}{\partial x}+ \frac{\partial p}{\partial \theta} \frac{\partial v}{\partial y}+\frac{\partial p}{\partial \theta} \frac{\partial}{\partial \theta}(\dot \theta)=0$

Now, following the book, by expanding the total derivative on the left I would get:

$\underbrace{\frac{\partial}{\partial t}(\frac{\partial p}{\partial \theta}) + u \frac{\partial }{\partial x}(\frac{\partial p}{\partial \theta}) + v \frac{\partial }{\partial y}(\frac{\partial p}{\partial \theta}) + \frac{d \theta}{dt}\cdot \frac{\partial }{\partial \theta}(\frac{\partial p}{\partial \theta})}_{\frac{d}{dt}(\frac{\partial p}{\partial \theta})} + \underbrace{\frac{\partial p}{\partial \theta} \frac{\partial u}{\partial x}+ \frac{\partial p}{\partial \theta} \frac{\partial v}{\partial y}}_{\frac{\partial p}{\partial \theta} \vec \nabla \cdot \vec V}+\frac{\partial p}{\partial \theta} \frac{\partial}{\partial \theta}(\dot \theta)=0$

Using divergence operator, this has 6 terms, however, in the final result given by the book

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I would miss the term

$\frac{d \theta }{d t} \cdot \frac{\partial }{\partial \theta}(\frac{\partial p}{\partial \theta})$

Why? I see no reason why this term is skipped.

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Potential temperature is conserved considering adiabatic processes and thus: $\text{d} \theta/\text{d}t = 0$. Otherwise the term needs to be retained.

On page 84 (just below figure 4.3) they say that given an adiabatic process, parcel flow is constrained to a surface of potential temperature, so I think they just assume adiabatic flow.

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    $\begingroup$ but then the last term should also vanish. $\endgroup$
    – MichaelW
    Commented Jun 22, 2023 at 14:26
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    $\begingroup$ I agree. Maybe someone else spots something? $\endgroup$ Commented Jun 22, 2023 at 15:02
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    $\begingroup$ could be an error in the book... (?) $\endgroup$
    – MichaelW
    Commented Jun 22, 2023 at 22:03
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    $\begingroup$ Tend to agree again :) $\endgroup$ Commented Jun 23, 2023 at 11:09

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