Earth Science Asked on May 22, 2021
I refer to this question, I asked recently in the physics board:
It doesn’t make sense to type it again, so I link to it.
Maybe the question can be answered here.
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: $$theta=Tleft(frac{P_0}{P}right)^{frac{R_d}{c_p}}tag{1}$$ 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 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 $frac{partialtheta}{partial z}=0$ 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, $frac{partial theta}{partial t}+ufrac{partial theta}{partial x}+vfrac{partial theta}{partial y}+wfrac{partial theta}{partial z}=0$ 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 $theta$ ($ufrac{partial theta}{partial x}+vfrac{partial theta}{partial y}=0$). 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.
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).
Correct answer by BarocliniCplusplus on May 22, 2021
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.
Answered by J. Fregin on May 22, 2021
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