<p>'Locally' and 'synoptically' are conflicting adjectives. Synoptic means overall, or large horizontal scale- think thousands of kilometers.</p>

<p>You are right on some accounts. Synoptically, pressure doesn't change much, depending on your qualitative definition of 'much.' On a percent scale or day-to-day experience, absolute temperature fluctuates more than pressure. But the gap of those fluctuations result in changes of density, leading to changes in buoyancy.</p>

<p>The vertical momentum equation :$$\frac{\partial w}{\partial t}+(\vec{v}\cdot\nabla_H)w+w\frac{\partial w}{\partial z}=-\frac{1}{\rho}\frac{\partial P}{\partial z}- g + \nabla^2w$$ when applied to synoptic scales can be treated as approximately hydrostatic: $$\frac{\partial P}{\partial z}=-\rho g$$.
From this equation, as well as the ideal gas law, the <a href="https://en.wikipedia.org/wiki/Hypsometric_equation" rel="nofollow noreferrer">hypsometric equation</a> may be derived.</p>

<p>Although the vertical gradient of pressure is large, gravity is also large, which explains why the atmosphere is not lost to space and vertical motions aren't usually large in horizontal scale.</p>

<p>Using said hypsometric equation, it can be shown that changes in vertically-averaged temperature lead to synoptic scale changes in surface pressure.</p>

CC BY-SA 3.0
2018-03-27T02:00:04.803Z
I am sort of getting it, but can you clarify.  Agreed, variations in pressure are relatively small: from what I can see 950 mb is in some cases the largest departure recorded in some countries, and in normal weather it looks like +/- 20 mb is big - about 2% from average.  Temperature is 270K +/-50K, so a 20% variation on the same day across the earth is not outrageous.  But if the sun shines brightly in a volcano caldera (so limited ability of air to flow in/out) when the air heats up doesn't that mean local pressure goes up (mostly) and the expansion upward slightly reduces density?
eSurfsnake
CC BY-SA 3.0
2018-03-27T13:31:20.550Z
Sure. Not only is pressure a function of temperature and density via the ideal gas law, but by a force balance, it can be shown that (from scale analysis of the first equation) the change in pressure with height is a function of density. If you want to go to a smaller scale, then the second equation is not really valid, and the first equation must be used. Often meteorologists like to consider isobaric surfaces on the synoptic scale, which changes the first equation such that $w$ turns into $\omega$ and $-\frac{1}{\rho}\frac{\partial P}{\partial z}$ turns into $-\frac{\partial gZ}{\partial p}$
BarocliniCplusplus
CC BY-SA 3.0
2018-03-27T13:43:27.967Z
Pressure is conserved on an isobaric surface, leading to the idea that density is inversely proportional to temperature. This allows meteorologists the ability to neglect the types of buoyancy effects * on a large scale. * On a smaller scale, isobaric surfaces may be used, but buoyancy cannot be neglected. Instead, the vertical momentum equation is derived with a buoyancy term, implying upward motion. .
BarocliniCplusplus
CC BY-SA 3.0
2018-03-28T00:38:08.977Z
thanks.  It is a challenge understanding the relative scale and balance issues if you have not had much practical experience, like me.
eSurfsnake
CC BY-SA 3.0
2018-03-28T12:28:06.533Z
No problem. For details one example on how scale analysis works, see: https://en.wikipedia.org/wiki/Scale_analysis_(mathematics)#Example:_vertical_momentum_in_synoptic-scale_meteorology. Some more information on scales: https://en.wikipedia.org/wiki/Meteorology#Spatial_scales.
BarocliniCplusplus
CC BY-SA 3.0
2018-03-29T03:50:04.873Z
One last question - I think I am close.  If there were rapid but short heating of a circle 100m in radius, one could imagine a parcel that is a short cylinder, 1m thick, that is 10K above ambient T.  Idealizing a lot, that cylinder would climb due to bouyancy.  It seems the key insight is that $\frac{dP}{dz}$ is much larger than $\frac{dT}{dz}$ since $P=p_0e^{-kz}$.  If this cylinder gets to the 500mb level, $P$ is roughly half what it was, and $T$ has dropped by maybe 7% or so.  Given $P=\rho RT$, and P adjusts rapidly, the cylinder grows vertically to something over 2m to balance everything?
eSurfsnake
CC BY-SA 3.0
2018-03-29T14:59:30.580Z
Often parcels aren't really given a real geometric shape, but instead, thought of as a differential volume (in dynamics, it may be thought as a differential cube, but it is understood that it won't stay a cube for long). Heating for your question would be neglected on the synoptic scale, where the hydrostatic equation is valid. For heating on the scale you want to be considered, Most of the Navier Stokes equation must be considered (https://www.ess.uci.edu/~yu/class/ess228/lecture.2.conservation.pdf).
BarocliniCplusplus
CC BY-SA 3.0
2018-03-29T15:06:45.247Z
It is worth noting that the $k$ is proportional to $\frac{1}{\bar{T}}$, where the bar is the mean. If you want to do a more explicit derivation, you can consider that approximate Taylor Polynomial $T=T_0 - \Gamma z$, where $\Gamma$ is the lapse rate. For a parcel undergoing dry adiabatic motions, $\Gamma=\Gamma_d \approx 0.0098 K m^{-1}$. I get the impression you are mostly interested in convection. You might want to look into potential temperature (https://en.wikipedia.org/wiki/Potential_temperature).
BarocliniCplusplus
CC BY-SA 3.0
2018-03-29T16:41:47.057Z
Convection is where I am starting since (1) it is observable where I live and (2) it matters a bit to my activities.  But I am an old electrical engineer with really solid math and physics, and recently took an interest in meteorology.  It is a tough topic to self-study, because there are so many nuances, scales, etc.  But I get the zen of what you are saying: imagine tiny differential blocks like one does in fluid mechanics or materials science.  The key where I am now is that P is what it is at each altitude and parcels take it on; T will drop, but slowly, $\rho$ is mostly P-driven.
eSurfsnake
CC BY-SA 3.0
2018-03-29T16:43:06.523Z
I am going to keep asking questions on this site, but if you want to do a few hours of paid work to guide me along, let me know.  The notes were very helpful.
eSurfsnake
CC BY-SA 3.0
2018-03-26T16:58:43.783Z
2018-03-27T13:29:40.640Z
2018-03-27T13:29:40.640Z
BarocliniCplusplus
<p>I have been learning meteorology a bit on my own, and one thing that never gets answered clearly is the relationship of pressure, temperature, and density.</p>

<p>When I looked it up, the density of air at various temperatures varies quite a bit - if I recall, from something like 1.1 to 1.5 kg/m<sup>3</sup>.  </p>

<p>My confusion comes from trying to understand weather formation.  A typical text says that, for instance, heating of the earth leads to thermals that rise due to their higher temperature <em>and lower density</em>, leading altitude increasing and to temperature dropping at the dry adiabatic lapse rate, then the moist adiabatic lapse rate, etc.  But it is always done using "parcel" arguments.</p>

<p>From basic thermodynamics, heating could lead to a combination of temperature increase, pressure increase, and physical expansion (if there is no fixed container).  </p>

<p>Specifically, the ideal gas law gives:</p>

<p>$$PV=nRT $$
or
$$P=\frac{n}{V}T$$ (ignoring the constant of R)
or
$$P=\rho T$$where $\rho$ equals density.</p>

<p>Am I wrong to assume that the simplifying assumption is that, to a first order approximation, the density can't really change much about an area since a parcel of air could expand, but the whole atmosphere locally can't (since the 'parcels' all abut)?  In other words, ground heating causes temperature to rise, and thus presumably pressure, but to a negligible degree decreases density?  </p>

<p>This would seem to imply higher pressure from heating actually leads to a net kinematic effect of driving the column of air upwards (as molecular pressure against the ground increases), where it can expand against lower pressure air above due to the fact pressure falls quickly with altitude - but there really isn't expansion or a decrease in density near the surface itself.</p>

<p>The typical "parcel lifted by a helicopter" argument fails to explain how it really all holds together and actually happens, since there is no 'container' for the gas in reality.</p>

<p>Another way to see this is in this example.  Suppose I have a hot tub in Antarctica in winter (I have one in Idaho in winter which is close enough).  The ambient temperature may be -30C or about 240K, and when I open the cover there is water at about 30C or 300K.  I can see thermals climb; the air, perhaps, is heated to 270K (being generous).  But the new air that gets sucked in and heated over the surface to 270K and then rises would immediately seem to be at higher pressure than ambient air since $$P=\rho T$$ I could even believe the kinematically more energetic air could cause an expanding 'plume' radially a bit as it rises, since the higher P and T slightly pushes out on less energetic ambient air.  But that just means $\Delta P$ is slighlty reduced to take into account the expanded volume of the plume; $\rho$ has dropped slightly, but nowhere near how fast it will drop if it gains altitude and has truly lower pressure about it.</p>

<p>In reality, it seems:</p>

<ol>
<li>locally the (synoptic) pressure is close to fixed - and varies only a few percent globally at sea level;</li>
<li>the gradient of pressure change is in the $z$ direction;</li>
<li>local heating increases T and thus P, except to the degree that increased P allows it to form an expanding plume (which is a few percent, so second order);</li>
<li>Really, the more energetic (kinetic) molecules have no where to go but up, what is really causing any updraft.  </li>
</ol>

CC BY-SA 3.0
2018-03-26T16:23:29.383Z
You can't neglect R
BarocliniCplusplus
<meteorology>
Relationship between pressure, temperature and density in meterology
eSurfsnake
# Relationship between pressure, temperature and density in meterology

(Asked by: username_0 on 2018-03-26)

I have been learning meteorology a bit on my own, and one thing that never gets answered clearly is the relationship of pressure, temperature, and density.

When I looked it up, the density of air at various temperatures varies quite a bit - if I recall, from something like 1.1 to 1.5 kg/m3.

My confusion comes from trying to understand weather formation. A typical text says that, for instance, heating of the earth leads to thermals that rise due to their higher temperature *and lower density*, leading altitude increasing and to temperature dropping at the dry adiabatic lapse rate, then the moist adiabatic lapse rate, etc. But it is always done using "parcel" arguments.

From basic thermodynamics, heating could lead to a combination of temperature increase, pressure increase, and physical expansion (if there is no fixed container).

Specifically, the ideal gas law gives:

$$PV=nRT $$
or
$$P=\frac{n}{V}T$$ (ignoring the constant of R)
or
$$P=\rho T$$where $\rho$ equals density.

Am I wrong to assume that the simplifying assumption is that, to a first order approximation, the density can't really change much about an area since a parcel of air could expand, but the whole atmosphere locally can't (since the 'parcels' all abut)? In other words, ground heating causes temperature to rise, and thus presumably pressure, but to a negligible degree decreases density?

This would seem to imply higher pressure from heating actually leads to a net kinematic effect of driving the column of air upwards (as molecular pressure against the ground increases), where it can expand against lower pressure air above due to the fact pressure falls quickly with altitude - but there really isn't expansion or a decrease in density near the surface itself.

The typical "parcel lifted by a helicopter" argument fails to explain how it really all holds together and actually happens, since there is no 'container' for the gas in reality.

Another way to see this is in this example. Suppose I have a hot tub in Antarctica in winter (I have one in Idaho in winter which is close enough). The ambient temperature may be -30C or about 240K, and when I open the cover there is water at about 30C or 300K. I can see thermals climb; the air, perhaps, is heated to 270K (being generous). But the new air that gets sucked in and heated over the surface to 270K and then rises would immediately seem to be at higher pressure than ambient air since $$P=\rho T$$ I could even believe the kinematically more energetic air could cause an expanding 'plume' radially a bit as it rises, since the higher P and T slightly pushes out on less energetic ambient air. But that just means $\Delta P$ is slighlty reduced to take into account the expanded volume of the plume; $\rho$ has dropped slightly, but nowhere near how fast it will drop if it gains altitude and has truly lower pressure about it.

In reality, it seems:

1. locally the (synoptic) pressure is close to fixed - and varies only a few percent globally at sea level;
2. the gradient of pressure change is in the $z$ direction;
3. local heating increases T and thus P, except to the degree that increased P allows it to form an expanding plume (which is a few percent, so second order);
4. Really, the more energetic (kinetic) molecules have no where to go but up, what is really causing any updraft.

**Tags:** meteorology

> **Comments:**
> **username_1**: You can't neglect R  

---

## 1 answers

### Answer by username_1 on 2018-03-26. Score: 1

'Locally' and 'synoptically' are conflicting adjectives. Synoptic means overall, or large horizontal scale- think thousands of kilometers.

You are right on some accounts. Synoptically, pressure doesn't change much, depending on your qualitative definition of 'much.' On a percent scale or day-to-day experience, absolute temperature fluctuates more than pressure. But the gap of those fluctuations result in changes of density, leading to changes in buoyancy.

The vertical momentum equation :$$\frac{\partial w}{\partial t}+(\vec{v}\cdot\nabla\_H)w+w\frac{\partial w}{\partial z}=-\frac{1}{\rho}\frac{\partial P}{\partial z}- g + \nabla^2w$$ when applied to synoptic scales can be treated as approximately hydrostatic: $$\frac{\partial P}{\partial z}=-\rho g$$.
From this equation, as well as the ideal gas law, the [hypsometric equation](https://en.wikipedia.org/wiki/Hypsometric_equation) may be derived.

Although the vertical gradient of pressure is large, gravity is also large, which explains why the atmosphere is not lost to space and vertical motions aren't usually large in horizontal scale.

Using said hypsometric equation, it can be shown that changes in vertically-averaged temperature lead to synoptic scale changes in surface pressure.

> **Comments:**
> **username_0**: I am sort of getting it, but can you clarify.  Agreed, variations in pressure are relatively small: from what I can see 950 mb is in some cases the largest departure recorded in some countries, and in normal weather it looks like +/- 20 mb is big - about 2% from average.  Temperature is 270K +/-50K, so a 20% variation on the same day across the earth is not outrageous.  But if the sun shines brightly in a volcano caldera (so limited ability of air to flow in/out) when the air heats up doesn't that mean local pressure goes up (mostly) and the expansion upward slightly reduces density?  
> **username_1**: Sure. Not only is pressure a function of temperature and density via the ideal gas law, but by a force balance, it can be shown that (from scale analysis of the first equation) the change in pressure with height is a function of density. If you want to go to a smaller scale, then the second equation is not really valid, and the first equation must be used. Often meteorologists like to consider isobaric surfaces on the synoptic scale, which changes the first equation such that $w$ turns into $\omega$ and $-\frac{1}{\rho}\frac{\partial P}{\partial z}$ turns into $-\frac{\partial gZ}{\partial p}$  
> **username_1**: Pressure is conserved on an isobaric surface, leading to the idea that density is inversely proportional to temperature. This allows meteorologists the ability to neglect the types of buoyancy effects * on a large scale. * On a smaller scale, isobaric surfaces may be used, but buoyancy cannot be neglected. Instead, the vertical momentum equation is derived with a buoyancy term, implying upward motion. .  
> **username_0**: thanks.  It is a challenge understanding the relative scale and balance issues if you have not had much practical experience, like me.  
> **username_1**: No problem. For details one example on how scale analysis works, see: https://en.wikipedia.org/wiki/Scale_analysis_(mathematics)#Example:_vertical_momentum_in_synoptic-scale_meteorology. Some more information on scales: https://en.wikipedia.org/wiki/Meteorology#Spatial_scales.  
> **username_0**: One last question - I think I am close.  If there were rapid but short heating of a circle 100m in radius, one could imagine a parcel that is a short cylinder, 1m thick, that is 10K above ambient T.  Idealizing a lot, that cylinder would climb due to bouyancy.  It seems the key insight is that $\frac{dP}{dz}$ is much larger than $\frac{dT}{dz}$ since $P=p_0e^{-kz}$.  If this cylinder gets to the 500mb level, $P$ is roughly half what it was, and $T$ has dropped by maybe 7% or so.  Given $P=\rho RT$, and P adjusts rapidly, the cylinder grows vertically to something over 2m to balance everything?  
> **username_1**: Often parcels aren't really given a real geometric shape, but instead, thought of as a differential volume (in dynamics, it may be thought as a differential cube, but it is understood that it won't stay a cube for long). Heating for your question would be neglected on the synoptic scale, where the hydrostatic equation is valid. For heating on the scale you want to be considered, Most of the Navier Stokes equation must be considered (https://www.ess.uci.edu/~yu/class/ess228/lecture.2.conservation.pdf).  
> **username_1**: It is worth noting that the $k$ is proportional to $\frac{1}{\bar{T}}$, where the bar is the mean. If you want to do a more explicit derivation, you can consider that approximate Taylor Polynomial $T=T_0 - \Gamma z$, where $\Gamma$ is the lapse rate. For a parcel undergoing dry adiabatic motions, $\Gamma=\Gamma_d \approx 0.0098 K m^{-1}$. I get the impression you are mostly interested in convection. You might want to look into potential temperature (https://en.wikipedia.org/wiki/Potential_temperature).  
> **username_0**: Convection is where I am starting since (1) it is observable where I live and (2) it matters a bit to my activities.  But I am an old electrical engineer with really solid math and physics, and recently took an interest in meteorology.  It is a tough topic to self-study, because there are so many nuances, scales, etc.  But I get the zen of what you are saying: imagine tiny differential blocks like one does in fluid mechanics or materials science.  The key where I am now is that P is what it is at each altitude and parcels take it on; T will drop, but slowly, $\rho$ is mostly P-driven.  
> **username_0**: I am going to keep asking questions on this site, but if you want to do a few hours of paid work to guide me along, let me know.  The notes were very helpful.  

---

