The question Why does the air pressure at the surface of the earth exactly equal the weight of the entire air column above it asks why the air pressure at any elevation is equal to the weight of the column of air above it in the atmosphere. I'm not convinced by any of the answers. Also, I'm going to slightly change the topic from a tall column of air in the atmosphere to the much simpler problem of a tall column of gas in a laboratory; the many processes that take place in the earth's atmosphere muddy the waters too much.

All of the answers boil down to considering a small parcel of gas; gravity pushes that parcel downwards, the gas-pressure differential pushes the parcel upwards, static equilibrium requires that the two forces be equal, and this results in the desired conclusion. Indeed, Feynmann (Chapter 40.1) makes this same argument in his Lectures, so I suppose it must be correct :-). Many physics textbooks make essentially the same argument for equilibrium water pressure, as well.

My problem is that a gas does not exist in small packets surrounded by thin weightless membranes, as those explanations all seem to imply. Gas molecules flow freely throughout space. I don't understand what it means for a "packet" of gas to be in equilibrium when molecules are entering and leaving the packet all of the time. Presumably the same number of molecules enter it as leave it, but the conditions for that to occur seem like they must be substantially more complex than the above explanations imply, especially when collisions are taken into account.

Is this another case where, because the mean free path of gas molecules is so small, they drift very slowly and hence behave as if they were really in small fixed packets? If so, would the above claims not apply in a less dense atmosphere (e.g., a high vacuum)?

Here are two explanations that seem plausible to me. Is either of them correct? Note that:

  • Again, I'm neglecting large-scale atmospheric effects such as solar radiation, so that the answers will be relevant to a column of gas in a laboratory.

  • Both explanations disregard collisions. However, a collision between two molecules obviously occurs only if the two molecules are at the same height. Since collisions are elastic and thus conserve total kinetic energy, the average kinetic energy of all molecules at any height is thus not affected by collisions. Since both proposals below are based largely on the concept of gravitational potential energy changing into kinetic energy as molecules fall, they should thus not be affected by collisions.

Here are the two theories:

  1. Gravity makes molecules gradually accelerate downwards. Neglecting collisions, the molecules closer to the earth would thus be (on average) moving faster. They are still moving in random directions, but are moving faster than the molecules higher up due to their lower gravitational potential energy. Similarly, because molecules tend to fall, there are more molecules lower down than higher up. Thus, lower heights will have a higher concentration of molecules, moving at higher speeds. This will result in a diffusion gradient of molecules moving upwards. This diffusion gradient will eventually balance the gravitational pull downwards, resulting in equilibrium. Result: close to the bottom we will have more molecules, and they will be moving faster, resulting in more kinetic energy per unit volume, which means more pressure. But while the argument seems to make sense qualitatively, that's a long way from a quantitative proof.
  2. Here's a completely different (and quantitative) argument. Consider dropping a single molecule from some height h. It will accelerate downwards until it bounces elastically into the earth and then slow down as it rises up to the original height h, and then do it over and over again. A bit of simple Newtonian mechanics quickly predicts the kinetic energy as a function of height, and then the velocity as a function of height, and then the amount of time spent in any small height interval, and then the fraction of the total time that the molecule spends in any small height interval. Now do this with many molecules that never collide. The number density at any height is then proportional to the fraction of time a molecule spends at that height. And the pressure at any height must be the number density times the kinetic energy per molecule. In fact, a few simple calculations (that I would be happy to include if requested) quickly result in precisely the conclusion that we wanted --- that the pressure change in any parcel is exactly equal to the weight of that parcel. It also predicts that temperature increases closer to the ground, due to the gravity-induced higher kinetic energy per molecule. Wonderful! But is it correct? Note that since molecules move more slowly at higher altitude, it also predicts that the number density increases at high altitude, which seems wrong.

I like theory #2 because it's quantitative and it exactly predicts that pressure at any height will equal the weight of the gas above it. Its prediction that number density increases at higher altitudes seems hard to stomach, but seems unavoidable. If molecules move more slowly at high altitudes (which they almost certainly do), then they spend more time traversing the same height delta at high altitudes than low, and they thus are statistically more likely to be found at high altitudes.

  • $\begingroup$ Temperature is not the same as higher kinetic energy per molecule. A block of steel thrown out into space does not decrease in temperature despite losing kinetic energy to gravity. I would recommend removing the bit about temperature from your question as that discussion is much more complicated. If interested in the answers, perhaps make a new question. $\endgroup$ – Rick Jan 8 '15 at 14:00
  • $\begingroup$ A block of steel thrown into space (in a graviational field) would gain bulk kinetic energy as it loses gravitational potential energy. No-one is saying that a steel block would lose temperature if it loses bulk kinetic energy. Temperature is a measure of the average microscopic kinetic energy of the molecules - it does not include bulk kinetic energy of the system itself. Bulk velocity is a relative concept: who is moving and who is not? Who is cold and who is hot? ;-) $\endgroup$ – Time4Tea Jan 8 '15 at 20:53

Ok, so David Hammen (rightly) debunked my previous answer, but I want to have another go because I feel that he hasn't provided a clear explanation of the molecular basis of hydrostatic pressure in a gas, which was the OP's original question:

I agree with David that, in a steady state, increased kinetic energy of the molecules is not the answer, as that would lead to a temperature gradient with the gas lower down having higher T; however, that's not possible because the gas column has to be in thermal equilibrium with its environment. (we're now considering a column of gas in a test tube in a lab, rather than the atmosphere, which complicates things).

However, that still left the question that, if a molecule of gas was to migrate from a higher point to a lower point in the column, it would have to lose gravitational potential energy. So, if that wasn't being converted to kinetic energy then where was it going?

Consider hypothetically that at some initial time (t=0), the molecules of gas are uniformly distributed across the height of the column and then they are suddenly released. Some gas molecules will move downwards under gravity and their kinetic energy will increase as they lose GPE. I think that has to happen, as there is nowhere else for the energy to go. So, there will be a temperature gradient but at this point the system is still transient - it hasn't yet reached steady-state. That extra heat energy will then dissipate to the walls of the container until the system is in thermal equilibrium (I'm assuming the initial downward motion of the molecules is significantly faster than the heat transfer).

So, in the steady state we have the situation that David described, where we have a pressure gradient, but the average kinetic energy of the molecules is the same throughout the column, because it is in thermal equilibrium. So, if increased kinetic energy isn't responsible for the hydrostatic pressure gradient, then the only other explanation is increased density of the molecules colliding with the walls of the container (i.e. there is a density gradient). I believe this density gradient is present due to the gravitational acceleration. If we consider an imaginary horizontal plane at some point in the column, there will be an increased likelihood of molecules travelling vertically downwards across the plane as opposed to going vertically upwards, because of the downwards acceleration. To conserve mass on average, in a steady-state this has to be balanced by there being more molecules per unit volume below the plane, which would increase the likelihood of molecules traversing the plane upwards (so the mass flux balances).

So, basically, at steady state, the hydrostatic pressure gradient is caused by a density gradient, which is present due to the gravitational acceleration. In this case, a molecule that migrates from a higher point to a lower one will still lose GPE and gain kinetic energy, but because it is steady-state, this will (on average) be balanced by another molecule going the other way (upwards), so the average temperature (and kinetic energy) will not increase.

  • $\begingroup$ The internal energy of an ideal gas depends linearly on temperature, not necessarily the average kinetic energy. In most problems the internal energy can be assumed to be its kinetic energy, and thus other forms of energy are neglected, but in this case the internal energy should include gravitational potential energy. $\endgroup$ – Rick Jan 8 '15 at 14:14
  • $\begingroup$ I don't think this is correct - at the molecular level, average kinetic energy of the molecules is directly related to temperature by the formula $E_k=\frac{3}{2}kT$. I don't believe its standard practice to include gravitational potential energy in internal energy. $\endgroup$ – Time4Tea Jan 8 '15 at 15:09
  • $\begingroup$ Its not standard practice, because typically the gravitation potential energy doesn't vary. Just as when you look at diatomic gasses you start with heat capacities based on 5 degrees of freedom, but then at extreme temperatures (like needed in scram jet calculations) the number of degrees of freedom goes up to 6 as the vibration of the bond becomes a viable energy storage mechanism. In most situations the gravitation potential energy doesn't vary enough to be considered. $\endgroup$ – Rick Jan 8 '15 at 15:54
  • $\begingroup$ If you'd like to read more about how other forms of energy are included in temperature here's a starting point. en.wikipedia.org/wiki/… $\endgroup$ – Rick Jan 8 '15 at 17:33
  • $\begingroup$ I don't see anything on that page that suggests that GPE is included in temperature - it seems to be referring to kinetic energy and modes of molecular vibration. Also, the first paragraph of the Wikipedia page on internal energy says this: In thermodynamics, the internal energy ... refers to energy contained within the system, while excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. $\endgroup$ – Time4Tea Jan 8 '15 at 19:41

Gravity makes molecules gradually accelerate downwards. Neglecting collisions, the molecules closer to the earth would thus be (on average) moving faster.

You cannot neglect collisions, at least not in the part of the atmosphere where the atmosphere acts like a gas. Collisions remain important until you get to the exobase. Above the exobase, the atmosphere is so rare that collisions can be ignored. Modeling the exosphere is messy because now you have to worry about solar flares, the Earth's magnetic field, and a varying distribution of components (e.g., the exosphere is dominated by hydrogen).

Collisions are extremely important in the thick part of the atmosphere (up to 120 km or so). At sea level, the mean free path (average distance between collisions) in the atmosphere is less than 1/10 of a micrometer. At 20 kilometers altitude, the mean free path is about one micrometer. By the time you get to 120 kilometers, the mean free path grows to about a meter. Collisions remain important until you get to the exobase.

Temperature in the atmosphere follows a complex profile. The highest temperatures are in the two highest layers of the atmosphere, the thermosphere and exosphere.

Your second argument suffers many of the same problems as your first. You cannot ignore collisions. Collisions are an essential part of what make a gas a gas. Things are a bit murky in the one part of the atmosphere, the exosphere, where collisions can be ignored. There is one easy way to model the exosphere: It's essentially a vacuum. Getting past that easy model is non-trivial. Even the thermosphere (the next layer down, where the space station orbits) is problematic to model. The atmosphere is still thin enough in the thermosphere that it doesn't act quite like a gas.

It's the mesosphere on down where the atmosphere acts like a gas. That the components are constant colliding with one another is what makes pressure in the lower atmosphere equal to the weight of all the stuff above. This isn't necessarily true in the upper atmosphere.

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    $\begingroup$ thanks for your answer. I've edited the original post to argue why the proposals it gives are robust in the face of collisions. Also, while it's nice to hear reasons why the proposed answers are wrong, it would also be nice to hear what the correct answers are :-). Reading between the lines, it sounds like you're saying that molecular collisions are what makes the pressure increase be equal to the weight of molecules above it. Can you make that more quantitative? Or was the original post correct in saying that this is indeed a very non-trivial question, and a quantitative answer would be hard? $\endgroup$ – JoelG Jan 7 '15 at 15:10
  • $\begingroup$ You already gave a reference to a correct answer in your question @JoelG, Feynman Volume 1, section 40.1. I similarly argued from a perspective of hydrostatic equilibrium in my answer to that other question. Hydrostatic equilibrium is the answer. You just don't like it. $\endgroup$ – David Hammen Jan 7 '15 at 16:08
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    $\begingroup$ I contest that the collisions have anything to do with why the pressure is equal to the weight of all the stuff above (divided by the area). The only reason this would not be true in the upper atmosphere is that pressure is an averaged quantity and the rarefied gas would make the fluctuations large, but if the pressure were taken over larger areas or time frames the pressure would still always be the weight of the stuff above (divided by the area it was above) $\endgroup$ – Rick Jan 8 '15 at 14:29
  • $\begingroup$ @Rick - Keep in mind that hydrostatic equilibrium is by far the scientifically accepted explanation. Hydrostatic equilibrium not only explains pressure variations in a gas, it also explains pressure variations in a liquid (e.g., the oceans), an elasto-plastic solid (e.g., the interior of the Earth), and a plasma (e.g., the interior of the Sun). $\endgroup$ – David Hammen Jan 8 '15 at 14:49
  • $\begingroup$ @David of course it is. And it's also in complete agreement with theory #2, which derives equations identical to those from hydro-static equilibrium, but doesn't require collisions (other than those on the pressure measuring device as without those there's no way to define pressure) $\endgroup$ – Rick Jan 8 '15 at 15:05

As David mentions you cannot neglect collisions and correctly predict the density profile. However, you can neglect them and still achieve the correct density profile. The reason that theory #2 seems to predict higher average densities at higher elevations is that it only considers one particle. Or more precisely, many particles with the exact same energy. If rather than all the particles bouncing back up to the same height, you had many particles bouncing at low heights and fewer particles bouncing at great heights, then each particle is more likely to be found at higher altitudes but the expected number of particles can be higher at lower altitudes. This could create an average density that decreased with height. In fact, you could create any arbitrary density profile (as long as you stayed away from the quantum limit). This is due to the fact that the density and temperature profiles depend on the collisions between the particles to equilibrate.

  • $\begingroup$ Good, that makes sense. $\endgroup$ – JoelG Jan 8 '15 at 22:25

UPDATE: I agree with David Hammen's reply that this answer is not correct, because in a steady-state there cannot be a thermal gradient in an enclosed column of gas (unless it is being heated externally somehow). Please see my more recent answer, which I believe provides the correct explanation of the molecular basis of hydrostatic pressure in a column of gas.

I completely agree with the answer provided by David Hammen, that you can't neglect collisions, as collisions between molecules are what transfers pressure between different points in a fluid (liquid or gas).

As to why the pressure is higher at the bottom of a column of gas: from the perspective of molecular thermodynamics, pressure is caused by particle collisions. If you have particles with higher energy or more particles impacting per unit area on a surface then you will have a higher apparent pressure. Particles at the bottom of the gas column have more kinetic energy because of the acceleration due to gravity and the fact that they have lost gravitational potential energy compared with the particles at the top of the column.

Think about it this way: if you just take one particle by itself and drop it from 1000m onto the ground, it will impact with more energy than if you dropped it from the same point onto a target just 1m below. So, effectively, it is creating more pressure after the 1000m drop, because it has more kinetic energy. So, I think, technically, even without the collisions, the pressure would still vary in the same way - the collisions are just a mechanism for transferring the kinetic energy from one point to another through a very dense cloud of particles.

So, basically, the pressure is higher because the particles at the bottom have more kinetic energy, having given up their gravitational potential energy. If one of those particles was to wriggle it's way back up to the top, it would have to gain gravitational potential energy to do that, so it would lose kinetic energy. I hope that helps!

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    $\begingroup$ This answer isn't correct. The kinetic energy of a gas and its temperature are closely related to one another (very simply related in an ideal gas). The highest temperatures in the Earth's atmosphere are in the thermosphere and exosphere, the two highest layers of the atmosphere. $\endgroup$ – David Hammen Jan 7 '15 at 14:11
  • $\begingroup$ Ok, that's a good point - I will think on it some more. $\endgroup$ – Time4Tea Jan 7 '15 at 15:11
  • $\begingroup$ Is the temperature profile of the Earth's atmosphere muddying the waters a bit unnecessarily? That profile depends on many complex factors (see question 40949 for details). The answer to this question here should be equally valid for a column of gas in a test tube. This answer seems to be saying much the same as the OP hypothesis #2 -- that it's simply the tradeoff between potential & kinetic energy. $\endgroup$ – JoelG Jan 7 '15 at 15:19
  • $\begingroup$ @JoelG - So lets unmuddy the waters and use a column of gas in a very tall (200 km should do it), thermally insulated test tube. After the system comes into equilibrium, what you'll find is a constant temperature throughout the test tube. That is the state that maximizes the entropy. Our atmosphere is far from that thermodynamic equilibrium because it is heated by sun (mostly from below) and radiates out into space. $\endgroup$ – David Hammen Jan 7 '15 at 15:47
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    $\begingroup$ @David Hammen: so is this where we're at now? 1. Everyone accepting that isotemperature is the equilibrium state, and that hydrostatic pressure behaves as desired. 2. No proof for iso-temperature, other than the reductio ad absurdum of noting that the lack of iso-temperature would result in perpetual motion (which I suppose is good enough). 3. No proof for the desired hydrostatic-pressure behavior on a molecular level. I'm not denying that it's correct, but like Time4Tea I think a molecular-level explanation would be nice. I guess "simple" and "statistical mechanics" are often at odds :-( $\endgroup$ – JoelG Jan 8 '15 at 3:09

Gravity induced thermal gradients in a gas column is not a new idea. It's probably of comparable age to the isothermal idea. Now many physicists (and others) have debated this for quite a while. We know what the majority opinion is. BUT... of course we are PROPER scientists, we respect the scientific method so instead of treating this like voting in an election ("Science is not a decomcracy" is a phrase you might have heard of) let's look at it without bias and stop falling into the "appeal to authority" logical trap.

EVIDENCE. Do we have any ? Well there is the atmosphere, the troposphere appears to have a pretty linear lapse rate, it is of the right sign (cool at top, hot at bottom). So it's a big gas column and looks like consistent with gravity induced gradient. (Score 1-0). Ah, but..but, all the other things winds, convection, GHE...OK..what about Venus. Again, from measurements, negative lapse rate=gravity gradient (Score 2-0)...Ah,but..but..CO2 thermal runaway..OK..all the other planets with enough data, shall we say (Score 4-0).

Yes, but these are all complicated systems, there are probably loads of things going on in all these planetary atmospheres which just coincidentally all make them look like a computable -g/Cp gravity induced temperature profile/lapse rate. Yeah, all of them, COINCIDENTALLY.

Never mind, lets get back to the lab. How about some real experiments with tubes of gas (and other stuff if you like e.g. water) and lets see if a vertical column of gas/water/any stuff you like will exhibit a vertical temperature profile, and measure it, and see if it has a value around what we would compute.

Well it has been done. It is early days yet, just a few experiments but they do seem consistent with the gravity explanation. I'll let readers do their own google (or whatever) research on this as there are a lot of comments,blogs,some papers and (as always) a lot of random rubbish and incorrect physics. Have a look and see what you think.

This is quite exciting new stuff generally (2007 or later) and deserves some consideration. Personally I think the evidence is building up towards the correct (scientific truth ?) being Gravity Induced Thermal Gradient and not Isothermal. The Laws of Thermodynamics are not neccesarily "wrong" if this is so but we might perhaps say they need a little tweaking as they have been worded a little sloppily and failed to specify all the relevant conditions. A situation not uncommon when new things are discovered in science.


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