<p>If we'd be living in a dry atmosphere your reasoning is indeed correct. Air would rise adiabatically and air would loose about 9.8 °C/km (dry adiabatic lapse rate). This means constant potential temperature. However, Earths atmosphere isn't dry. As soon as a rising, moist air parcel reaches saturation, it will rise with a moist adiabatic lapse rate (6-7 °C/km). Potential temperature will rise (because if the parcel falls down again it will be warmer at the same level - see below). This causes clouds and sometimes rain and thus, reduces the overall water content in an air parcel.</p>
<p>If such a parcel is now going to fall again it will gain temperature according to the dry adiabatic lapse rate (because it lost water during ascension). Lets say initially a parcel starts rising with moist adiabatic lapse rate with temperature <span class="math-container">$T(t=0)$</span> and then falls down again and reaches the same level at <span class="math-container">$t=1$</span>. You will then observe that <span class="math-container">$T(t=0)&lt; T(t=1)$</span>. Thus, potential temperature must have increased as the parcel was rising.</p>

CC BY-SA 4.0
2020-09-08T15:36:04.960Z
And that is what I am answering. I'm not sure where my answer has flaws. If you have suggestions I'll gladly include them.
Joscha Fregin
CC BY-SA 4.0
2020-09-09T20:37:20.610Z
yes but my "problem" was a wrong conclusion (see my comment above).
MichaelW
CC BY-SA 4.0
2020-09-08T13:57:27.563Z
2020-09-08T15:47:38.420Z
2020-09-08T15:47:38.420Z
Joscha Fregin
<p>You may be forgetting that pressure also decreases with height (exponentially). Also, because <span class="math-container">$P=\rho R T$</span>, <span class="math-container">$\frac{dP}{dT}=\rho R$</span> (that is, <span class="math-container">$c_p$</span> does not appear). But I digress in answering your question.</p>
<p>Let's break down why potential temperature increases with height. Let's start with the equation: <span class="math-container">$$\theta=T\left(\frac{P_0}{P}\right)^{\frac{R_d}{c_p}}\tag{1}$$</span> Now, if we want to ask why <span class="math-container">$\theta$</span> increases with height, let's rephrase <span class="math-container">$(1)$</span> as a function of height:  <span class="math-container">$$\theta(z)=T(z)\left[\frac{P_0}{P(z)}\right]^{\frac{R_d}{c_p}}\tag{2}$$</span> Now if you follow <a href="https://earthscience.stackexchange.com/questions/19517/derivative-of-exner-function/19519#19519">my derivation</a> for finding <span class="math-container">$$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}\left(\frac{\partial T}{\partial z}+\frac{g}{c_p}\right)$$</span></p>
<p>You may note that the <span class="math-container">$\frac{\partial\theta}{\partial z}=0$</span> iff. <span class="math-container">$\frac{\partial T}{\partial z}=-\frac{g}{c_p}=-9.8 \textrm{ K km}^{-1}$</span> (dry adiabatic lapse rate). If <span class="math-container">$\frac{\partial T}{\partial z}&gt;-\frac{g}{c_p}$</span> then <span class="math-container">$\frac{\partial \theta}{\partial z}&gt;0$</span>.</p>
<p>This does not mean that the parcel is not adiabatic, though. For a parcel to be truly adiabatic, <span class="math-container">$\frac{\partial \theta}{\partial t}+u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}+w\frac{\partial \theta}{\partial z}=0$</span> Therefore, the only time a certain location can be considered adiabatic and have <span class="math-container">$\frac{\partial \theta}{\partial z}=0$</span> is when there is no heating (<span class="math-container">$\frac{\partial \theta}{\partial t}=0$</span>), and no horizontal advection of <span class="math-container">$\theta$</span> (<span class="math-container">$u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}=0$</span>). Note, I did not exclude vertical advection of <span class="math-container">$\theta$</span>, because that is implied in the condition that <span class="math-container">$\frac{\partial \theta}{\partial z}=0$</span>.</p>
<p>I'll also take pause here and list a few diabatic processes, and why there may be some justification to ignore them for your mental image.</p>
<ul>
<li>Radiation: with the exception of greenhouse gases, the atmosphere is mostly transparent.</li>
<li>Surface heating/ turbulent transport: more important near the surface (note that <span class="math-container">$\theta$</span> is a minimum at the surface).</li>
<li>Clouds/microphysical processes: discrete and are never everywhere all the time and moisture is a completely separate variable that varies all the time. There is a way around that (to some extent- the <a href="https://en.wikipedia.org/wiki/Equivalent_potential_temperature#:%7E:text=%2C%20is%20a%20quantity%20that%20is,condenses%20during%20that%20pressure%20change." rel="nofollow noreferrer">equivalent potential temperature</a>, for latent heat exchange).</li>
<li>Chemical processes: See below about how I remember <span class="math-container">$\theta$</span> increases with height.</li>
</ul>
<hr>
<p>If you want a good reminder of why <span class="math-container">$\theta$</span> increases with height, it might be easier to remember that  <span class="math-container">$sgn \left( \frac{\partial \theta}{\partial z}\right)$</span> is a good indicator of static stability. And thunderstorms need to stop at some point, even if that is the tropopause (where <span class="math-container">$\frac{\partial \theta}{\partial z}&gt;0$</span> due to ozone heating).</p>

CC BY-SA 4.0
2020-09-09T20:14:55.853Z
While reading I found my problem: I confused **active** transport of a particular parcel from z to z+Δz with two parcels 1, 2, **spatially** being separated by Δz at the same time. In the former case, assuming adiabatic state change, θ will not change while T changes by -g/c_p*Δz. However, for the second case there is no limiting restriction regarding θ1 and θ2, since 2 could be energetically more "charged" as compared to 1. (of course I know, that a positive gradient of θ(z) yields a statically stable layer structure, while  negative layered atmospheric structures are intrinsically unstable.)
MichaelW
CC BY-SA 4.0
2020-09-08T15:51:44.277Z
2020-09-08T15:51:44.277Z
BarocliniCplusplus
<p>I refer to this question, I asked recently in the physics board:</p>
<p><a href="https://physics.stackexchange.com/questions/578308/doesnt-increase-of-potential-temperature-with-height-contradict-adiabatic-natur">https://physics.stackexchange.com/questions/578308/doesnt-increase-of-potential-temperature-with-height-contradict-adiabatic-natur</a></p>
<p>It doesn't make sense to type it again, so I link to it.
Maybe the question can be answered here.</p>

CC BY-SA 4.0
2020-09-08T13:31:10.300Z
These slides are about adiabatic processes. Taking such adiabatic process, potential Temperature should be constant.
MichaelW
CC BY-SA 4.0
2020-09-10T08:58:26.893Z
Please [edit] your question so that it also contains all necessary information *here* (https://earthscience.meta.stackexchange.com/questions/1874/cross-site-duplicate-question-should-contain-all-information)
Jan Doggen
<atmosphere><temperature><atmosphere-modelling><troposphere>
Doesn't Increase of Potential Temperature with Height contradict Adiabatic Nature of Processes within Troposphere?
MichaelW
# Doesn't Increase of Potential Temperature with Height contradict Adiabatic Nature of Processes within Troposphere?

(Asked by: username_0 on 2020-09-08)

I refer to this question, I asked recently in the physics board:

<https://physics.stackexchange.com/questions/578308/doesnt-increase-of-potential-temperature-with-height-contradict-adiabatic-natur>

It doesn't make sense to type it again, so I link to it.
Maybe the question can be answered here.

**Tags:** atmosphere, temperature, atmosphere-modelling, troposphere

> **Comments:**
> **username_0**: These slides are about adiabatic processes. Taking such adiabatic process, potential Temperature should be constant.  
> **username_1**: Please [edit] your question so that it also contains all necessary information *here* (https://earthscience.meta.stackexchange.com/questions/1874/cross-site-duplicate-question-should-contain-all-information)  

---

## 2 answers

### Answer by username_3 on 2020-09-08. Score: 4 | Accepted answer

You may be forgetting that pressure also decreases with height (exponentially). Also, because $P=\rho R T$, $\frac{dP}{dT}=\rho R$ (that is, $c\_p$ does not appear). But I digress in answering your question.

Let's break down why potential temperature increases with height. Let's start with the equation: $$\theta=T\left(\frac{P\_0}{P}\right)^{\frac{R\_d}{c\_p}}\tag{1}$$ Now, if we want to ask why $\theta$ increases with height, let's rephrase $(1)$ as a function of height: $$\theta(z)=T(z)\left[\frac{P\_0}{P(z)}\right]^{\frac{R\_d}{c\_p}}\tag{2}$$ Now if you follow [my derivation](https://earthscience.stackexchange.com/questions/19517/derivative-of-exner-function/19519#19519) for finding $$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}\left(\frac{\partial T}{\partial z}+\frac{g}{c\_p}\right)$$

You may note that the $\frac{\partial\theta}{\partial z}=0$ iff. $\frac{\partial T}{\partial z}=-\frac{g}{c\_p}=-9.8 \textrm{ K km}^{-1}$ (dry adiabatic lapse rate). If $\frac{\partial T}{\partial z}>-\frac{g}{c\_p}$ then $\frac{\partial \theta}{\partial z}>0$.

This does not mean that the parcel is not adiabatic, though. For a parcel to be truly adiabatic, $\frac{\partial \theta}{\partial t}+u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}+w\frac{\partial \theta}{\partial z}=0$ Therefore, the only time a certain location can be considered adiabatic and have $\frac{\partial \theta}{\partial z}=0$ is when there is no heating ($\frac{\partial \theta}{\partial t}=0$), and no horizontal advection of $\theta$ ($u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}=0$). Note, I did not exclude vertical advection of $\theta$, because that is implied in the condition that $\frac{\partial \theta}{\partial z}=0$.

I'll also take pause here and list a few diabatic processes, and why there may be some justification to ignore them for your mental image.

* Radiation: with the exception of greenhouse gases, the atmosphere is mostly transparent.
* Surface heating/ turbulent transport: more important near the surface (note that $\theta$ is a minimum at the surface).
* Clouds/microphysical processes: discrete and are never everywhere all the time and moisture is a completely separate variable that varies all the time. There is a way around that (to some extent- the [equivalent potential temperature](https://en.wikipedia.org/wiki/Equivalent_potential_temperature#:%7E:text=%2C%20is%20a%20quantity%20that%20is,condenses%20during%20that%20pressure%20change.), for latent heat exchange).
* Chemical processes: See below about how I remember $\theta$ increases with height.

---

If you want a good reminder of why $\theta$ increases with height, it might be easier to remember that $sgn \left( \frac{\partial \theta}{\partial z}\right)$ is a good indicator of static stability. And thunderstorms need to stop at some point, even if that is the tropopause (where $\frac{\partial \theta}{\partial z}>0$ due to ozone heating).

> **Comments:**
> **username_0**: While reading I found my problem: I confused **active** transport of a particular parcel from z to z+Δz with two parcels 1, 2, **spatially** being separated by Δz at the same time. In the former case, assuming adiabatic state change, θ will not change while T changes by -g/c_p*Δz. However, for the second case there is no limiting restriction regarding θ1 and θ2, since 2 could be energetically more "charged" as compared to 1. (of course I know, that a positive gradient of θ(z) yields a statically stable layer structure, while  negative layered atmospheric structures are intrinsically unstable.)  

---

### Answer by username_2 on 2020-09-08. Score: 4

If we'd be living in a dry atmosphere your reasoning is indeed correct. Air would rise adiabatically and air would loose about 9.8 °C/km (dry adiabatic lapse rate). This means constant potential temperature. However, Earths atmosphere isn't dry. As soon as a rising, moist air parcel reaches saturation, it will rise with a moist adiabatic lapse rate (6-7 °C/km). Potential temperature will rise (because if the parcel falls down again it will be warmer at the same level - see below). This causes clouds and sometimes rain and thus, reduces the overall water content in an air parcel.

If such a parcel is now going to fall again it will gain temperature according to the dry adiabatic lapse rate (because it lost water during ascension). Lets say initially a parcel starts rising with moist adiabatic lapse rate with temperature $T(t=0)$ and then falls down again and reaches the same level at $t=1$. You will then observe that $T(t=0)< T(t=1)$. Thus, potential temperature must have increased as the parcel was rising.

> **Comments:**
> **username_2**: And that is what I am answering. I'm not sure where my answer has flaws. If you have suggestions I'll gladly include them.  
> **username_0**: yes but my "problem" was a wrong conclusion (see my comment above).  

---

