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

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?

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

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,

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

• Side note: I think I may have gotten some minus signs mixed up. But the spirit of the derivation stays the same. Let's not correct them, in case some teacher decides to make this an exam question..... Kinda kidding about the last part. Kinda. – BarocliniCplusplus May 31 '20 at 1:17
• Nice answer. Given the detail of your answer I think it's worthy to note that you used the chain rule in (6) and $d/dx 1/f(x) = - 1/f(x)^2 d/dx f(x)$. – J. Fregin Jun 1 '20 at 0:18
• @J.Fregin Thanks! When I learned calculus, I never could keep straight whether the divisor or the dividend was subtracted within the quotient rule. So I realized that the quotient rule was effectively the chain rule. To get to (6), you need to take $\frac{\partial}{\partial z} (3)$ – BarocliniCplusplus Jun 1 '20 at 12:09