<p><span class="math-container">$$\theta=T\left(\frac{P_0}{P}\right)^\frac{R_d}{c_p} \tag{1}$$</span>
<span class="math-container">$$\alpha=R_d T P^{-1} \tag{2}$$</span>
<span class="math-container">$$\frac{\partial P}{\partial z}=-\rho g \tag{3}$$</span></p>

<p><a href="https://earthscience.stackexchange.com/questions/19517/derivative-of-exner-function/19519#19519">It can be shown that <span class="math-container">$$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}(\frac{\partial T}{\partial z}+\frac{g}{c_p }) \tag{4}$$</span></a></p>

<p>Therefore <span class="math-container">$$\frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}(\frac{\partial T}{\partial z}+\frac{g}{c_p })\tag{5}$$</span></p>

<p>So if we take <span class="math-container">$\frac{\partial \alpha}{\partial z}$</span>, we find <span class="math-container">$$\frac{\partial \alpha}{\partial z}=R_dP^{-1}\frac{\partial T}{\partial z}-R_dTP^{-2}\frac{\partial P}{\partial z}\tag{6}$$</span></p>

<p>Then we can rewrite (6) using (2):</p>

<p><span class="math-container">$$\frac{\partial \alpha}{\partial z}=\frac{\alpha}{T}\frac{\partial T}{\partial z}-\alpha P^{-1}\frac{\partial P}{\partial z}\tag{7}$$</span></p>

<p>Divide by <span class="math-container">$\alpha$</span> <span class="math-container">$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}- P^{-1}\frac{\partial P}{\partial z}\tag{8}$$</span></p>

<p>Employing (3) <span class="math-container">$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ P^{-1}\rho g\tag{9}$$</span></p>

<p>Which is the equivalent of </p>

<p>Employing (3) <span class="math-container">$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ P^{-1}\alpha^{-1} g\tag{10}$$</span></p>

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

<p><span class="math-container">$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ \frac{g}{H-H_0-P\alpha}\tag{11}$$</span></p>

<p>Since enthalpy is conserved, <span class="math-container">$H=H_0$</span></p>

<p><span class="math-container">$$\frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ \frac{g}{c_p T}\tag{12}$$</span></p>

<p>And finally,</p>

<p><span class="math-container">$$\therefore \frac{1}{\alpha}\frac{\partial \alpha}{\partial z}=\frac{1}{T}\left(\frac{\partial \alpha}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+ \frac{g}{c_p}\right)=\frac{1}{\theta}\frac{\partial \theta}{\partial z}$$</span></p>

CC BY-SA 4.0
2020-05-31T01:17:21.677Z
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
CC BY-SA 4.0
2020-06-01T00:18:30.467Z
Nice answer. Given the detail of your answer I think it's worthy to note that you used the chain rule in (6) and $d/dx 1/f(x) = - 1/f(x)^2 d/dx f(x)$.
Joscha Fregin
CC BY-SA 4.0
2020-06-01T12:09:24.153Z
@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
CC BY-SA 4.0
2020-05-31T01:12:45.117Z
2020-05-31T01:12:45.117Z
BarocliniCplusplus
<p>While studying about atmospheric waves, I found a passage in my book in which a equation involved the term  <span class="math-container">$$ \frac{1}{\alpha }\frac{\partial \alpha }{\partial z}$$</span>
where <span class="math-container">$\alpha$</span> is the specific volume of air , which was replaced by <span class="math-container">$$ \frac{1}{\theta }\frac{\partial \theta }{\partial z}$$</span>, where <span class="math-container">$\theta$</span> 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?</p>

<meteorology><thermodynamics>
What is the relationship between the derivatives of specific volume and potential temperature with height?
paulo
# What is the relationship between the derivatives of specific volume and potential temperature with height?

(Asked by: username_0 on 2020-05-30)

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?

**Tags:** meteorology, thermodynamics


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## 1 answers

### Answer by username_1 on 2020-05-31. Score: 3

$$\theta=T\left(\frac{P\_0}{P}\right)^\frac{R\_d}{c\_p} \tag{1}$$
$$\alpha=R\_d T P^{-1} \tag{2}$$
$$\frac{\partial P}{\partial z}=-\rho g \tag{3}$$

[It can be shown that $$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}(\frac{\partial T}{\partial z}+\frac{g}{c\_p }) \tag{4}$$](https://earthscience.stackexchange.com/questions/19517/derivative-of-exner-function/19519#19519)

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,

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

> **Comments:**
> **username_1**: 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.  
> **username_2**: Nice answer. Given the detail of your answer I think it's worthy to note that you used the chain rule in (6) and $d/dx 1/f(x) = - 1/f(x)^2 d/dx f(x)$.  
> **username_1**: @J.username_2 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 $\frac{\partial}{\partial z} (3)$  

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