I don't understand how to calculate the surface pressure in high/low pressure systems. For example, let's say we have a planet without oceans whose surface is completely flat. Then at every point of the surface pressure should be total atmosphere mass * g / surface area. Now, when we add heterogeneous warming from the sun, I know that raising warm air should leave behind low pressure at the surface. My confusion arises from the fact that the mass of the column of air above surface hasn't changed after the warm air rose, so surface pressure should change either. The only way I could explain this is if we advect the warm air away but without first having a pressure gradient that generates wind I dont see that happening.

How can I calculate the pressure at surface created by warm air rising and cold air sinking? Is there a formula for eulerian pressure/mass budget?


1 Answer 1


I found the answer to this question. When air is warmed up, it expands. Surface pressure stays the same initially but at any given altitude the pressure in the warmer column of air will be lower then corresponding pressure at the same altitude in a colder column of air. This causes air to move from the warm column to the cold column at high altitudes (because of pressure gradient) reducing the surface pressure of the warm column and increasing the one of the colder column because mass was moved away.

An illustration of this phenomenon is given at page 342 of chapter 11 of Practical Meteorology (https://www.eoas.ubc.ca/books/Practical_Meteorology/)

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Notation: H = high pressure perturbation, L = low; black dots represents air parcels; thin arrows are winds.

  1. (i) Initial conditions: two air columns A and B with equal mass are given with the same temperature profile
  2. (ii) Temperatures change: the column A becomes colder and shrinks (for example irradiating heat back into space at night) and the column B becomes colder and expands (for example warmed up by absorbed radiation from the sun during the day). The surface pressure is still the same because the columns have the same mass. Only the total volume of the column has changed.
  3. (iii) Pressure gradient creates horizontal winds: when measuring the pressure P at height z > 0 for column A and column B, we will get that $P_A(z) < P_B(z)$. This drives a horizontal wind (stronger at higher altitudes) by accelerating air over time: $\frac{dV_i}{dt} = -\frac{1}{\rho} \frac{dP}{di}$, where $V$ is the wind speed in the ith direction (so replace with x, y) - this is described in chapter 10 of afore mentioned book.
  4. (iv) Winds redistribute mass between the air columns: air will move from high pressure points to low pressure points, so from column B to column A. The density change over time is $\frac{d\rho}{dt} = \sum\limits_{i} \rho \frac{dV_i}{dt}$. More mass in colder column A means that surface pressure is higher $p_A(0) > p_B(0)$ whilst at higher altitude, due to lower volume of colder air, we will still have $p_A(z) < p_B(z)$. So we will have horizontal winds from warm columns of air to colder columns of air at higher altitudes, and surface winds from colder columns of air to warmer columns of air. Since volume is conserved, the volume outflows at higher latitude in the warmer column must equate the volume inflows at lower latitude (and viceversa for the colder column of air) which means air also moves up in the warmer column and down colder column. This is exactly how a Hadley cell works!

Note this is still a simplified scenario but it does a good job at illustrating the dynamics of the creation of high/low pressure points and circulation cells. Other important considerations are discussed in chapter 10.

  • $\begingroup$ I would like some equations... but it seems this could work in flat geometry... But what about spherical planet? Is planet uniformally heated or only on one side? $\endgroup$
    – Vid
    Commented Jan 10, 2023 at 23:36
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jan 11, 2023 at 0:11
  • $\begingroup$ @Vid I have been trying to summarize the equations but it's quite complex and it will require me still some time. My reference for this phenomenon is eoas.ubc.ca/books/Practical_Meteorology Chapter 11 (general circulation) page 342. The equations are described throughout chapter 10. $\endgroup$
    – Redirectk
    Commented Jan 24, 2023 at 21:21

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