What is the relationship between the derivatives of specific volume and potential temperature with height?

Question

While studying about atmospheric waves, I found a passage in my book in which a equation involved the term $$ \frac{1}{\alpha }\frac{\partial \alpha }{\partial z}$$ where $\alpha$ is the specific volume of air , which was replaced by $$ \frac{1}{\theta }\frac{\partial \theta }{\partial z}$$, where $\theta$ is the potential temperature, but with no details on how this was done. I tried to find a relationship between specific volume and potential temperature that would give me this equivalence, but with no success. How can I prove this?

Answer 1

$$\theta=T\left(\frac{P_0}{P}\right)^\frac{R_d}{c_p} \tag{1}$$ $$\alpha=R_d T P^{-1} \tag{2}$$ $$\frac{\partial P}{\partial z}=-\rho g \tag{3}$$

It can be shown that $$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}(\frac{\partial T}{\partial z}+\frac{g}{c_p }) \tag{4}$$

Therefore $$\frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}(\frac{\partial T}{\partial z}+\frac{g}{c_p })\tag{5}$$

So if we take $\frac{\partial \alpha}{\partial z}$, we find $$\frac{\partial \alpha}{\partial z}=R_dP^{-1}\frac{\partial T}{\partial z}-R_dTP^{-2}\frac{\partial P}{\partial z}\tag{6}$$

Then we can rewrite (6) using (2):

$$\frac{\partial \alpha}{\partial z}=\frac{\alpha}{T}\frac{\partial T}{\partial z}-\alpha P^{-1}\frac{\partial P}{\partial z}\tag{7}$$

Divide by $\alpha$ $$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}- P^{-1}\frac{\partial P}{\partial z}\tag{8}$$

Employing (3) $$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ P^{-1}\rho g\tag{9}$$

Which is the equivalent of

Employing (3) $$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ P^{-1}\alpha^{-1} g\tag{10}$$

Now, it may get a little sketchy in this derivation. But we're close. Assuming an adiabatic transformation ($dH=c_p dT-Pd\alpha=0$ and $H-H_0=c_pT-P\alpha$), we can rewrite (10) to look a little funky

$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ \frac{g}{H-H_0-P\alpha}\tag{11}$$

Since enthalpy is conserved, $H=H_0$

$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ \frac{g}{c_p T}\tag{12}$$

And finally,

$$\therefore \frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\left(\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ \frac{g}{c_p}\right)=\frac{1}{\theta}\frac{\partial \theta}{\partial z}$$