Does hot air rise -- from a statistical-mechanical viewpoint

Question #6329 asks whether and why hot air rises. The consensus answer is straightforward: - hot air is less dense than cold air - therefore, by the principal of buoyancy, it rises. However, it is a thermodynamical explanation rather than a statistical-mechanical one. I would like to ask the same question from a statistical-mechanical point of view.

Furthermore, while the buoyancy argument makes sense to me for an enclosed volume of hot air, it does not make sense to me for an unenclosed volume.

For an enclosed volume H of hot air, the statistical-mechanical explanation for it rising seems straightforward. As the air in an enclosed volume gets hotter, $PV=nRT$ says that it expands but its pressure lowers. The air below H is thus at a lower temperature but higher pressure than H. Thus, the average kinetic energy per unit volume is higher below H than in H. Thus, if there were a barrier below H, more impulse would be transferred to the barrier from below than from above. The same would be true for the barrier above H; but since atmospheric pressure is lower at higher altitude, the energy transferred would be less. Thus the container would rise.

However, for our unenclosed-gas example, there is no such barrier -- and a non-existent barrier cannot absorb any impulse! So now let's reconsider the problem, but this time without a barrier surrounding the hot region H.

Here's my explanation. Please poke holes into it :-)

Before we heated it, the air in the region of space H had the same temperature and pressure as the surrounding air. Then, some agent added energy to H to heat up those molecules. Now the molecules in H are hotter than their surroundings. However, the number density of molecules in H has not changed -- i.e., we are not yet at equilibrium.

With the molecules in H suddenly moving faster (and, as usual with gases, moving in random directions), all of the boundaries of H will experience a diffusion gradient. At every boundary, more molecules will leave H than enter. To be clear, these boundaries of H have no actual physical barrier of any sort; they are merely theoretical boundaries.

The energetic molecules in H thus start to leave H in all directions. Up, down and sideways. They are replaced by a smaller number of less energetic molecules from outside. As that happens, the number density of molecules in H diminishes. At some point, we reach thermal equilibrium -- but with most of the heat energy having left the region H.

So where has it gone? The air below H was originally warmer and at a higher pressure than the air above H. Hence, the diffusion gradient at the boundary above H will be steeper than the gradient at the boundary below H. Hence, more heat energy will flow upwards than downwards. However, there will definitely be heat energy that moves both downwards and sideways from H.

Conclusion: while hot air does tend to generally move up, it also moves downwards and sideways too.

Is this a correct conclusion and reasoning?

  • $\begingroup$ I'm interested to see other ideas on this. Buoyancy is a pretty slow process that happens over long distances, while stat. mech. deals with fast things (speed of sound-ish) over small distances (mean free path). The question makes me think of lattice Boltzmann methods. As I understand it (vaguely), the can recover continuum behavior of fluids from a discrete statistical formulation. $\endgroup$ Commented Sep 15, 2014 at 18:25
  • $\begingroup$ The general answer is, no, statistical mechanics is not a good tool for dealing with mesoscopic and macroscopic non-equilibrium processes. Actually, it's not even a good tool for most of thermodynamics, because the calculation of thermodynamic quantities form first principles is computationally enormously demanding. $\endgroup$
    – CuriousOne
    Commented Sep 15, 2014 at 21:01

1 Answer 1


Your answer would be correct only for high vacuum - a density so low that collisions between molecules can be ignored. At atmospheric pressure the mean free path of an air molecule is only about a micron. If the heated volume is much larger than this we can ignore diffusion and thermal conduction in the short run. The constant collisions create what is effectively a barrier between the hot and cold air, stopping them from diffusing into each other.

Edit. A steep temperature gradient forms between the hot and the cold. Its width is small, but still larger than the mean free path. Each molecule is surrounded by molecules of a similar energy.

It's true that there isn't a physically distinct barrier. Nevertheless, I can draw an imaginary thin interface zone. The pressure on one boundary of this zone is due to molecules arriving from the hot side. Most will be scattered back into the hot side by molecules of approximately the same energy, and this defines the pressure. Similarly on the cold side. If the pressures are not equal then the boundary zone, which has very little mass, will accelerate.

(Note that there is a little wrinkle here, which also arises for a solid membrane. The pressure calculated from the temperature and density only applies strictly if the surface of the membrane is at the same temperature as the gas.)

Consequently the pressure on either side of the boundary must be equal in mechanical equilibrium, and when it isn't, the gas will accelerate. This is why the air will first expand after being suddenly heated, because it has a higher pressure.

Edit. The gas can expand rapidly when there is a significant pressure difference - up to the speed of sound, which is of the same order as the mean molecular speed. Under these circumstances the gas has a bulk velocity - in other words the distribution of molecular velocities is not isotropic, but a majority are moving in one direction. Each molecule is struck more often by others travelling in that direction.

Once the average pressure is equalized, the gravitational pressure gradient is less in the lower density gas, so the pressure still cannot be equal on both side of the whole boundary. This causes the hot low density air to rise.

Once mechanical equilibrium is established, the slower processes of diffusion and thermal conduction will eventually cause the temperature and density to equilibrate. In practice these are greatly accelerated by turbulence, which increases the surface area of the boundary.

Edit. Everything above is relatively straightforward: the gas behaves as a continuous fluid. The concept of a continuous fluid becomes much more blurred when there is a shock (also known as a shock wave or shock front). Strictly speaking this occurs in your example, because the gas is heated suddenly. The thickness of the shock cannot be calculated from continuum mechanics, but is a few mean free paths. Moreover it is inevitably dissipative, as mechanical work is turned into heat and heat conducts rapidly over very short distances. As long as the gradients are kept much wider than the mean free path, shocks can be avoided and the continuum approach is OK.

  • $\begingroup$ Thanks, this is starting to make more sense. But still not completely. I'm editing the response to ask clarification questions. $\endgroup$
    – JoelG
    Commented Sep 16, 2014 at 13:19

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