"Heat rises" or "warm air rises" is a widely used phrase (and widely accepted phenomenon).

Does hot air really rise? Or is it simply displaced by colder (denser) air pulled down by gravity?

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    $\begingroup$ What is the difference? $\endgroup$ – Mark Eichenlaub Mar 4 '11 at 1:38
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    $\begingroup$ I suppose I'm looking for clarification. It strikes me that in the absence of cold(er) air, the only relevant force acting on the hot air would be gravity, pulling it down. Is there another force that I'm missing? $\endgroup$ – jasonmklug Mar 4 '11 at 1:42
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    $\begingroup$ Yes, WSC... that's about the gist of what I'm wondering. If the cold air were stationary (maybe we assume the cold air is magically unaffected by gravity), would the hot air still rise? $\endgroup$ – jasonmklug Mar 4 '11 at 1:45
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    $\begingroup$ @Distil: In the absence of other fluid around it, it simply disperses (drifts off into the vacuum) or sit there tightly confines (micro bubbles in some solid). The question only makes sense in the context of a bulk fluid, and then the two cases are one and the same. $\endgroup$ – dmckee --- ex-moderator kitten Mar 4 '11 at 2:32
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    $\begingroup$ "Does hot air really rise? Or is it simply displaced by colder (denser) air pulled down by gravity?" ... maybe the question is worded prone to misinterpretation, but the fragment "by colder (denser) air" does give a clear indication that the connection between those properties is recognized. Any nitpicking about the difference without showing what difference it would make to the process in question, is not very helpful. $\endgroup$ – Gyro Gearloose Feb 14 '16 at 17:21

16 Answers 16


The mechanism responsible for the rising of hot air is flotation: Hot air is less dense than cold air and hence air pressure will exert an upwards force, in the same way air rises in water. Now if cold air was magically unaffected by gravity, then it would not be able to exert pressure on the hot air and thus it would not rise.

The statement that "heat rises", by the way, is not universally true. Look at water. Here, it is the cold water that is less dense than warm water (at least in the temperature regime of importance to freezing). In winter, when water gets colder, the cold water raises to the top and eventually will freeze, while the water below remains liquid for the moment.

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    $\begingroup$ "heat rises" is generally nonsense! "Heat" is not a stuff, and diffuses. But fighting this wording is like fighting windmills. $\endgroup$ – Georg Mar 4 '11 at 10:15
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    $\begingroup$ Agreed. The problem is that heat in everyday-language isn't the same as the physical concept of heat. Same with energy, work, order, theory. What can one do... $\endgroup$ – Lagerbaer Mar 5 '11 at 18:07
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    $\begingroup$ With respect to water, the temperature at which warm water becomes less dense than cold is 4° C. van.physics.illinois.edu/qa/listing.php?id=1736 $\endgroup$ – Ben Hocking Jun 22 '11 at 18:15

Buoy, does it ever. To contrast the previous answers I will give a mathematical description and a concrete example to bolster the intuitive understanding.

Ideal Gas Law

From thermodynamics we know that pressure, $P$, temperature $T$, and density $\rho$ (or specific volume $v=1/\rho$) are related through an equation of state. For suitable gases (including air at atmospheric conditions) this equation is the ideal gas law:

\begin{equation} \tag{1} \label{igas} P = \rho R T \end{equation}

where $R$ is the specific gas constant, which can be determined by the chemical makeup of the gas under consideration (e.g. $R_{air}=287.058 \:\mathrm{J kg^{−1} K^{−1}} $).


As already mentioned by Helder Velez, Archemides' Principle informs us that an object immersed in a fluid will experience an upward force equal to the weight of the displaced fluid, where 'up' is the direction of decreasing density gradient.1 Mathematically, this may be stated as:

\begin{equation} \tag{2} \label{buoy} \mathbf{F_b} = -\rho V \mathbf{g} \end{equation}

where $\mathbf{g} = -g\mathbf{\hat{k}}$ is the body force vector (usually gravity).

Air Bubble in water

Consider a small air bubble, initially at rest near the bottom of a pool, at thermal equilibrium (same temperature) as the pool water. The buoyancy force acting on the bubble is given by equation \ref{buoy}, and the weight of the bubble is given by $\mathbf{F_g} = m\mathbf{g}$. The subscript $w$ refers to water and the subscript $a$ refers to the air in the bubble. Applying Newton's second law yields:

\begin{align} m_a \mathbf{a} &= \sum \mathbf{F} \\ m_a \mathbf{a} &= \mathbf{F_g} + \mathbf{F_b} \\ m_a \left( a_x \mathbf{\hat{i}} + a_y \mathbf{\hat{j}} + a_z \mathbf{\hat{k}} \right) &= -m_a g\mathbf{\hat{k}} + \rho_w V_b g\mathbf{\hat{k}} \\ m_a a_z &= \rho_w V_a g - m_a g \\ a_z &= g\left(\frac{\rho_w V_a}{m_a} - 1\right) \\ a_z &= g\left(\frac{\rho_w V_a}{\rho_a V_a} - 1\right) \\ a_z &= g\left(\frac{\rho_w}{\rho_a} - 1\right) \\ \end{align}

where I have used $m_a = \rho_a V_a$. Here, it can be seen that the bubble will accelerate upward whenever $\rho_w > \rho_a$. Leveraging the fact the pressure varies linearly with depth in a static fluid, you can prove to yourself that $\rho_w \gg \rho_a$ for bubbles in most pools.


Parcel of Air

Now consider a similar scenario, where instead of a pool we have a room full of air at uniform temperature $T_\infty$ and the bubble is now a parcel of air which has been heated to a slightly elevated temperature $T_\infty + \Delta T$. I will use the subscripts $c$ for the air in the cool room, and $h$ for the air within the hot parcel.

If we perform a similar analysis to the bubble in the pool, we will go through the same motions as the derivation above, and end with a similar expression for the initial acceleration of the hot parcel:

\begin{equation} a_z = g\left(\frac{\rho_c}{\rho_h} - 1\right) \end{equation}

In this case however, we can use equation \ref{igas} to further simplify the result:

\begin{align} a_z &= g\left(\frac{P/(R_{air} T_\infty)}{P/\left(R_{air} \left[T_\infty + \Delta T\right]\right)} - 1\right) \\ a_z &= g\left(\frac{T_\infty + \Delta T}{T_\infty} - 1\right) \\ a_z &= g\left(\frac{\Delta T}{T_\infty}\right) \\ \end{align}

I can think of no better mathematical affirmation of the adage Hot air rises than the above equation. Wherever $\Delta T > 0$, $a_z$ will be also. Conversely a cooler parcel will fall: $\Delta T < 0 \rightarrow a_z < 0$.


You might wonder:

Why is it that the bubble in the pool is so straightforward, yet the air parcel rising is not immediately obvious?

Three reasons come to mind:

  1. The bubble is well defined. It has a clear spherical boundary which is more or less maintained during its' ascent. On the other hand, our parcel is not visible, and even if it is spherical initially, it can stretch and morph at the mercy of any local air currents.
  2. The ratio $\rho_w/\rho_a$ is usually much bigger than $\rho_c/\rho_w$, making the motion of the bubble much more pronounced than that of the hot air parcel.
  3. The air parcel is subject to heat transfer. Imagine we wrapped our little air parcel in a tiny balloon. Even if its shape is maintained, the air parcel will transfer heat to the surrounding air as it rises, the temperature will drop so that $a_z \rightarrow 0$, and viscous forces will slow it to a halt.

Also note: the magnitude of acceleration is independent of the pressure. Whether we are in a pressure chamber at $10 \:\mathrm{atm}$ or on mount everest at $0.333 \:\mathrm{atm}$ it will always divide through.

Finally, I will point out that, even though the ideal gas law gives us a very elegant expression for acceleration, all other fluids (which I can think of) have equations of state with negative correlations between $T$ and $\rho$, meaning that a fluid parcel with an elevated temperature relative to a quiescent fluid of the same thermodynamic makeup will always have a buoyancy force of greater magnitude than its weight.

1For hydrostatic fluids and many flows the pressure gradient $\nabla P$ is nearly always aligned with the density gradient $\nabla \rho$. Specifically the direction of the body force vector $\mathbf{g}$ is opposite the density gradient.


Heat does only 1 thing in a closed system, and that is evenly distribute itself about the system as it reaches thermodynamic equilibrium. I dont think this is what you asking about though. I assume you are talking about hot air (hot being a relative term just meaning it is hotter then the surrounding air). This hot air will be less dense then the surrounding air, and will therefore want to be above the more dense, colder air. If you want to actualy see this, get a beaker of water and add some oil, this is same thing that happens with air (as both cases involve 2 liquids of different densities)

To answer the question exactlty, hot air does rise, and it is also displaced by cold air (though often from the side, not directly above it). And yes, gravity is the reason less dense liquids like to sit on top of more dense liquids


The statement "hot air rises" is not in general true, although often used.


Less dense air rises

Now usually, locally heated air will expand (because pressure will be similar to the pressure of the surrounding air) according to the universal gas law $PV=nRT$, and less dense air will experience buoyancy from the surrounding more-dense (cooler) air. Hot air will not rise if it's surrounded by hotter air...

Look at the example of a helium balloon, for example. Although the "air" inside the balloon might be colder than the surrounding air, it can still rise - because the gas inside is less dense. And if you created a thin-walled container with low-pressure air (80-20 mixture of nitrogen and oxygen), it could conceivably rise although it's at the same temperature as the surrounding air.

Look also at the air we breathe out: it contains oxygen, carbon dioxide, nitrogen and water. Now carbon dioxide has a higher molecular mass than oxygen, but the addition of water tends to lower the density of the air. So when a politician talks (produces "hot air"), the breath they produce may go up or down. It depends on the temperature of the surrounding air (if the air around him is warmer, for example because he's in a sauna, then the expired air will be cooler than the surrounding air; it may also have lower relative humidity and more carbon dioxide - so it will definitely sink). At sufficiently high relative humidity, in air close to body temperature,(a hot muggy day), it is possible that "hot air sinks".

We can do the math: the composition of air is roughly

in   out  mass  
80%  75%    28  nitrogen
20%  14%    32  oxygen
 0%   6%    18  water
 0%   4%    44  CO2
 1%   1%    40  argon

That makes the average molar mass for inspired air 28.9 g/mol, and 29.8 g/mol for expired air, using the most extreme case of dry air. We can calculate the relative temperatures at which these have the same density:

$$T_1 m_1 = T_2 m_2$$

Using the above numbers, if the temperature of the expired air is 37 °C (330 K), it has the same density as dry atmospheric air with a temperature of 28 °C. This means that when the surrounding air is hotter than 28 C, expired air ("hot air") will sink, even if the relative humidity is zero. It's hard to be a good politician when the airconditioning is broken...

So it's density, not temperature, than matters. Although one often implies the other.

  • $\begingroup$ This is much better than the accepted answer, IMO. I feel like I would enjoy most anything you write! $\endgroup$ – iconoclast Feb 8 '19 at 17:49

I have a master's in meteorology, so I can help you out. I apologize if this isn't as professional as some of the other responses, but I'm a little tired right now.

Just take, for example, this equation:

$$F_B = \left( \frac{\rho_0-\rho}{\rho}\right)\approx g\cdot\left( \frac{T-T_0}{T_0}\right)$$

We know that warm air has a lower density than cooler air. So if you want to prove to yourself that warm air has a greater buoyant force than cool air, just plug some numbers in. Just assume that rho-not has a value of 1.25 and that rho has a value of 1.00. That gives you a buoyancy of 0.25. Now, take some cooler air. Increase rho-not to around 1.15. This gives you a buoyancy force of 0.09. So indeed, warm air is more buoyant than cool air and thus experiences a positive buoyancy and rises.

Just keep in mind, though, that this is only valid for parcel theory. Obviously in the real world, there are more things going on than just this equation, but this should at least give you a basic understanding.


I'm going to go over your question bit by bit. Explain some of the language of your question, and then analyze the final answer. My explanations assume prior knowledge of the atomic reality of gases but little else.

First off "heat rises" is a term that should be avoided in a physics discussion. The term "heat" is referring to the transfer of thermal energy from one place to another. It is not a state-quantity. For instance, state quantities are things that are qualities of the matter it self. For instance mass is a state quantity. So is charge. These are the same regardless of other place and time. While "heat" is a description of change, not a description of state. We say a pan on the stove heated up. Or better still a flow of heat from the flame into the pan caused the pan to have a higher temperature. If we said the pan on the stove has heat, that is incorrect, the pan on the stove has thermal energy (a mesure of the mass and temperature of the object), and a temperature.

Reminder: temperature is a mesure of the average kinetic energy of a substance.

Rephrase: Does hot (greater temperature) air rise? Or is it displaced by cold (lower temperature) air?

First: Why does anything fall and rise in a gravitational field? Well it must have a force pushing it up. To change its potential energy (U=mg) a force must act on it.

What is the force that causes a fluid or gas to rise and fall? In all cases it can be described as a pressure.

Pressure is always a relative thing, this is because it isn't pressure that causes things to rise and fall it is a pressure difference or gradient. So what is important is the net pressure, or pressure difference.

First of all this is an important point. If the pressure in a volume is all the same: nothing changes. No air moves (besides individual particles that will move due to brownian motion).

So how can I create a pressure difference to cause one bit of air to rise? To be pushed up?

1) The easiest way is to control how packed the air is, its density. A greater packed group of molecules will have more atoms in a smaller space so it if each molecule is moving at the same speed more collisions occur between the edge of its volume (these changes in momentum cause a force) and it will exert more force: greater pressure.

2) But how do we mesure how fast the particles are going in volume of something? Because if the atoms move faster then there will be greater changes in momentum and more force. Temperature is the mesure of this, the average kinetic energy describes in essence how fast the particles are going.

What does 1) and 2) tell us? Well pressure is controlled by the speed of the particles, and how many of them are in the space. In thermodynamics the equation PV=nRT is used. R is a constant. n is the number of mol (a measure of the number of particles). This says the pressure and volume (V) are related to temperature (speed) and the amount (n).

This says that a hotter volume of the same substance will need to expand to maintain its outward pressure. A colder thing will contract. This is the process of hotter and colder liquids and gases becoming less dense or more dense.

FINAL (Q and A): A: Does hotter air rise? B: Or does cold air displace the hot air causing it to rise?

Well let's test the fist one, hotter implies that it is hotter then something. So if it is hotter then the air around it, the air will expand, the pressure will decrease (since PV is constant) and high pressure, lower density air will push it up: displacing it. Here we see the issue with the question: both A and B are true. If B were not, and cold air wasn't available then there would be no difference in densities, no difference in pressures, and nothing would change. You can not have A without B, and B without A, and mostly this is because a pressure gradient is necessary for changes to occur.

Could you have two gases where the hotter of the two was on the bottom? Yes. A light gas like helium, less dense because its molecules hate each other (personified; sorry), will float all the way out of the earth's atmosphere, leaving hot desert air below.


Another way to think about it is to look at how pressure changes with height. If we place a high density box shaped parcel of fluid immediately next to a lower density fluid parcel, the hydrostatic pressure gradient is greater in the former parcel. So, if say the average pressure of the two parcels is the same, the denser one will have higher pressure at the bottom than the lighter one, and lower pressure than the lighter one at the parcel tops. So the denser parcel will tend to push in at the bottom, and be displaced at the top. To a first order approximation the will rotate, trying to put the lighter parcel on top. If the fluid parcels are of identical composition then the warmer one will be lighter. Of course we a temperature regime in water, where the density versus temperature curve runs backwars between roughly 0C to 4C, and ice is lighter still. But in general warmer fluid is lighter.

In anycase the original question is rhetorical. Do we take the mental shortcut and think in terms of bounancy as a lifting force, or try to be more precise and consider the fluids interaction as the cause. In most cases, I'd prefer the former methodology, as that makes it easier to formulate the dynamics.


The Action force is simultaneous with the Reaction force. One can not happen without the other. Archimedes settled that a less dense fluid move on top of a denser fluid (see Buoyancy - Archimedes' principle) .
(and vice-versa: a denser..moves..to bottom..)

The Rayleigh-Taylor instability describes the evolution of the interface between the two layers. The atomic bomb mushroom cap is due to this effect. atomic bomb mushroom cap(quoting Wikipedia):

RT..is an instability of an interface between two fluids of different densities, which occurs when the lighter fluid is pushing the heavier fluid. This is the case with an interstellar cloud and shock system. The equivalent situation occurs when gravity is acting on two fluids of different density — with the dense fluid above a fluid of lesser density — such as water balancing on light oil.


Here is the mechanism, you can yourself figure out what happens -

There are two things at work here - gravity, and gas pressure. In the beginning, due to gravity, denser air is at bottom and lighter air is at the top. You may ask why it is this way to begin with and the answer is in the word "denser". At every level, there is an equilibrium density in the beginning and it is higher at the bottom and lower at the top. When some air near bottom is heated, it does not push up, all it does, is expands in all directions due to its increased temperature (and so increased pressure). Due to expansion, its density goes lower. And due to this lowering of density, the density equilibrium is disturbed. Then gravity brings back the density equilibrium by pulling dense air down more than it pulls the hot air. So hot air only expands, gravity does the rest.

Now you may ask why gravity pulls denser air more than the lighter air. Because denser air has more mass per volume and so more gravitational force per unit of volume. (GMm/(r*r)).

Therefore in actuality, even though it is dense air that is pushing down (due to gravity), the hot, less dense air has no where to go except up, and so it appears it is pushing up but it really does not (or we can say it pushes in all directions, not only up, due to its increased pressure). The cold air moves down only from sides, it can not move down from directly above because of increased pressure of hot air.

Similar question asked on this site - "Why does hot air rise in a column instead of cold air pressing down?"


"Hot air" is just air molecules (M) moving faster (F); "cold air" is M moving more slowly (S). The collisions between the FMs and SMs force both Ms in all directions (SM faster than before, FMs slower than before, but still faster than most SMs). The space below them, however, is crowded with SMs, so those FMs knocked upward keep going fast -- until they hit the (albeit fewer) SMs above them, continuing the process. Shake a half-full bag of popcorn: the big kernels work their way up, leaving the smaller and denser corn below.

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    $\begingroup$ "The space below them, however, is crowded with SM" No, the space below is lower, so is hotter, so there are FM. That was a error of you? $\endgroup$ – Santropedro Jan 7 '17 at 16:54

Just one word for you, the convection. Have you heard about it?

Convective heat transfer, often referred to simply as convection, is the transfer of heat from one place to another by the movement of fluids. Convection is usually the dominant form of heat transfer(convection) in liquids and gases

We are just too stupid to understand what it means, thinking that this is just a clever word for the nerds. How do you think hot masses are transferred if hot does not goes up and cool does not goes down?

Moreover, if warm air is not raising then what do you have, where does it go? You say that cool air is sucked into the hot place. What do you have then? Is all the air accumulated in one place in the space? I think that the density is overly high already when you heat the spot. That is why hot molecules start spreading out. This makes air less dense and, thus, lighter, and your mushroom raises up, as your child hydrogen baloon does. Yes, baloons physically raise up because they are lighter than the rest of the air and it is more efficient for the nature to have lighter objects at higher altitudes and have heavier masses at lower altitudes (nature does potential energy minimization).


Hot air is less dense than cold air. The particles have more energy and by bouncing into each other (faster) they cause the average space between themselves to be greater than that for cold air.

To ask, does a single hot particle rise above a single cold particle doesn't make sense. If they're in a vacuum, or even near each other in a vacuum, they will both be pulled by gravity to the bottom of that vacuum. In a sense, the vacuum is the "hottest" air, or least dense.

Back to the question of why a bunch of hot air will rise above cold air: It weighs less. The particles individually, of course, weigh the same. But if you take equal volumes of hot air and cold air, there will be MORE AIR in the cold volume. So it will settle to the bottom, because it there is a greater force on it. The hot air has nowhere else to be, but higher, or rising.

Of course, this neglects to address the fact that there will be a transfer of energy between hot and cold air-- but that gets to be beyond the scope of the question and ultimately, regardless of which air is hot, it is that air which will be less dense and sit on top of the fluid. In exactly the same way that a beaker of two fluids with unequal densities causes the less dense fluid to rise and sit on the top.


Maybe I missed it in the other answers however I haven't seen anyone mention the one constant of gravity.

Space doesn't pull things From the earth, the earth's gravity Pulls things towards it.

So however true that hotter air May rise and colder air May fall, it's the gravity that initiates the fall as the first cause. Then the first effect is that colder, denser air fills its container first from the ground up, pushing hotter air Up!


if you have a sauna (room full of hot air) and you inject a smaller quantity (say one gallon) of colder air through a pipe in the ceiling, the colder air will seem to "fall" down.

i think of it as "higher density volumes get pulled 'harder'/'faster' and leave the less dense volumes 'behind'" (in the context of the earth's gravity)


I am not sure that I am covering something already mentioned but I would like to underline an important idea: Hot-air balloons rise because there is a membrane around a less-dense substance and the balloon as a unit weighs less than an equal volume of air.

But hot air that is unconfined is a different situation. If you have a volume of air at one temperature and you introduce, in the simplest case, a single air molecule that is relatively energetic then it is more likely that it will end up higher in the volume of air as it bounces around because the original volume of air is more dense at the bottom and less dense at the top (due to gravity) and so there are more molecules to bounce off of at lower height.

But the rate of rise of a hot air molecules must be harder to predict than the rise of such molecules confined in a membrane and they rise for a different reason. Unconfined collection of hot molecules do not weigh less that the same number of cooler molecules and while they tend to occupy a greater volume and so in some sense are less dense, that is not why the hot air rises in a column of cooler air.


More correctly, warm air rises from the heat it receives from the Earths surface, even inside. Then it gets too high to receive heat and sinks, while at the same time, the cool air that has just recieved enough energy to rise takes its place.


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