I realised, reading another Phys.SE question about balloons moving forwards in an accelerating car that I don't really understand how buoyancy works. Particularly concerning, for a SCUBA diver.

The top answers to that question seem to claim that balloons get their "sense of down" from a pressure differential. They continue: when a car accelerates, the air at the back of the car becomes more dense, and at the front less dense, changing the plane of the pressure differential and so also, the balloon's sense of up. I find that extremely hard to credit. However, I realised that I don't really know why less dense things float in more dense things.

I'm fairly sure it's something to do with displacement of heavier things by lighter things, and I think pressure acting on the lighter thing's surface has something to do with it, but that's about it.

  • 1
    $\begingroup$ Wikipedia also suggests buoyancy is due to the pressure differential of the supporting fluid, unless I misunderstand. This just doesn't make sense, because the upwards force is proportional to the mass displaced, not the (much smaller) change in density of the supporting fluid across the object. $\endgroup$
    – Benjohn
    Commented Jul 2, 2014 at 21:21
  • $\begingroup$ And there is essentially no change in density in water, although, there is a change in pressure. $\endgroup$
    – Benjohn
    Commented Jul 2, 2014 at 21:33
  • $\begingroup$ Hmm. Although, the idea of the weight of the column pushing on an element of the top of the displacer, and the effective weight pushing up on an element of its bottom does make sense. Which is, essentially, a pressure differential. I think I've argued myself around :-) $\endgroup$
    – Benjohn
    Commented Jul 2, 2014 at 21:45
  • 1
    $\begingroup$ Right, I was reading your question which basically makes sense, but in your first comment I lost track of what it is you have a problem with in the first place :) By the way, this visualization is nice (click to enable Buoyancy visualization): phet.colorado.edu/sims/density-and-buoyancy/buoyancy_en.html $\endgroup$
    – BjornW
    Commented Jul 2, 2014 at 21:49
  • $\begingroup$ Heh, that's great @BjornW! You can work out the volume of the scales :-) $\endgroup$
    – Benjohn
    Commented Jul 2, 2014 at 21:53

5 Answers 5


Basic idea

Picture in your mind a deep ocean of water. Imagine a column of the water, going from the surface down to a depth $d$. That column of water has some weight $W$. Therefore, there is a downward force of magnitude $W$ on that column of water. However, you know the column of water is not accelerating, so there must be an upward force of magnitude $W$ pushing on that column. The only thing underneath the column is more water. Therefore, the water at depth $d$ must be pushing up with force $W$. This is the essence of buoyancy. Now let's do details.


The weight $W$ of a column of water of cross-sectional area $A$ and height $d$ is

$$W(d) = A d \rho_{\text{water}}$$

where $\rho_{\text{water}}$ is the density of water. This means that the pressure of water at depth $d$ is

$$P(d) = W(d)/A = d \rho_{\text{water}}.$$

Now suppose you put an object with cross sectional area $A$ and height $h$ in the water. There are three forces on that object:

  1. $W$: The object's own weight.
  2. $F_{\text{above}}$: The force of the water above the object.
  3. $F_{\text{below}}$: The force of the water below the object.

Suppose the bottom of the object is at depth $d$. Then the top of the object is at depth $d-h$. Using our results from before, we have

$$F_{\text{below}} = P(d)A=d \rho_{\text{water}} A $$


If the object is in equilibrium, it is not accelerating, so all of the forces must balance:

$\begin{eqnarray} W + F_{\text{above}} &=& F_{\text{below}} \\ W + (d-h) \rho_{\text{water}} A &=& d \rho_{\text{water}} A \\ W &=& h A \rho_{\text{water}} \\ W &=& V \rho_{\text{water}} \end{eqnarray}$

where in the last line we defined the object's volume as $V\equiv h A$. This says that the condition for equilibrium is that the weight of the object must be equal to its volume times the density of water. In other words, the object must displace an amount of water which has the same weight as the object. This is the usual law of buoyancy.

From this description I believe you can extend to the case of air instead of water, and horizontal instead of vertical pressure gradient.


I think pressure acting on the lighter thing's surface has something to do with it, but that's about it.

This is actually the beginning and the end of the WHOLE story. This, in theory is everything you need to know about buoyancy. Let's see how this statement plays out, and how it leads to the other pieces of knowledge you have gleaned about buoyancy.

You simply imagine a free body diagram for the floating / immersed body. The only forces on it are the pressure, everywhere normal to the body's surface, and the body's weight.

The nett force on the body from the surrounding fluid is then:

$$\mathbf{F} = \int_S\,p(\mathbf{r})\, \mathbf{\hat{n}}(\mathbf{r})\,\mathrm{d} S\tag{1}$$

where we sum up the pressure forces $p(\mathbf{r})\,\mathbf{\hat{n}}(\mathbf{r})$ acting on the elements of area $\mathrm{d} S$ in the direction of the unit normal $\mathbf{\hat{n}}(\mathbf{r})$ as a function of position $\mathbf{r}$ over the interface surface $S$ between the fluid and the body. That's all there is to it. Of course, it is hard to see from (1) alone what will happen to a body steeped in fluid, so let's move on to more practical answers.

We do a little trick: it turns out that you can always assume for buoyancy problems that the surface $S$ in (1) is a closed boundary of a volume (that's even when you deal with problems like boats that, ideally, are not totally submerged and the closed boundary would at first sight seem inapplicable). We first form the inner product of $\mathbf{F}$ with an arbitrary unit vector $\mathbf{\hat{u}}$ and then, given the closed surface, we can apply the divergence theorem to (1) for the volume $V$ within the closed surface $S=\partial\,V$:

$$\langle\mathbf{F},\,\mathbf{\hat{u}}\rangle=\oint_{\partial V}\,p(\mathbf{r})\,\mathbf{\hat{u}}\cdot\mathbf{\hat{n}}(\mathbf{r})\,\mathrm{d}S=\int_V\boldsymbol{\nabla}\cdot(p(\mathbf{r})\,\mathbf{\hat{u}})\,\mathrm{d}V=\mathbf{\hat{u}}\cdot\int_V\boldsymbol{\nabla}(p(\mathbf{r}))\,\mathrm{d}V$$

which, given the unit vector $\mathbf{\hat{u}}$ is arbitrary, means:

$$\mathbf{F} = \int_V\boldsymbol{\nabla}(p(\mathbf{r}))\,\mathrm{d}V\tag{2}$$

and we are to imagine the pressure field $p(\mathbf{r})$ that would be present in the fluid within the surface if the fluid weren't being displaced by the body taking up the volume $V$. From (2) we can immediately see the second piece of knowledge that you have heard of:

balloons get their "sense of down" from a pressure differential. [bold mine]

that is, there is no nett buoyant force on the body unless the pressure $p$ varies from place to place. Otherwise, $\boldsymbol{\nabla}(p(\mathbf{r}))$ is identically nought.

If you are not fully comfortable with the divergence theorem, think of and analyze a submerged cube. In a fluid where pressure does not vary with position , the force on each face is exactly balanced by the opposite force on the opposite face. Another case that gives intuition is a sphere in a fluid with a constant pressure everywhere: the force on any point is precisely balanced by the opposite force on the antipodal point. The divergence theorem argument simply lets you infer the generalness of conclusions like this that you can make for symmetrical objects.

Now let's move on to a pressure field that will be wonted to you as a scuba diver; taking the $\mathbf{\hat{z}}$ direction as down, the pressure field within a still fluid lying on the surface of a planet of radius much greater than depths we need to consider is:

$$p(\mathbf{r}) = (p_0+\rho\,g\,z)\,\mathbf{\hat{z}}\tag{3}$$

where $\rho$ is the fluid density, $g$ the gravitational acceleration and $p_0$ the pressure at $z=0$. If we plug this into (2) we get:

$$\mathbf{F} = \rho\,g\,\mathbf{\hat{z}}\,\int_V\,\mathrm{d}V = \rho\,g\,V_f\,\mathbf{\hat{z}}\tag{4}$$

where $V_f$ is the volume of fluid displaced. This is, of course, Archimedes' principle; it holds for regions of fluid small enough that the pressure variation is a linear function of position. Although it seems to be saying the "displaced fluid pushes back" as many vague explanations of buoyancy state, but this is nonsense. The displaced fluid isn't even there: the principle is merely the results of applying mathematical tricks to translate the fundamental principle, which is embodied in the your text that I quoted in the first line of this answer and in (1) and the "displaced fluid pushback" merely a mnemonic to recall the principle.

Two further comments are in order:

  1. Firstly, note that the answer in (4) is independent of $p_0$, Therefore, if the body isn't wholly submerged (like a working boat hull), then we can simply take the intersection of the volume with the fluid to be the volume $V$; the intersection of the fluid's surface with the volume then bounds the reduced volume and the force contribution on the top face is then nought (since we can arbitrarily set $p_0=0$ without changing our results).
  2. Secondly, again, if you are uncomfortable with the divergence theorem, do the analysis for a cube with its edges vertical and horizontal as a clarifying example. Although the pressure force varies across the vertical surfaces, pressure surfaces on each vertical face are still exactly opposed by those on the opposite face. The nett force is the difference between the force on the cube's bottom and top faces, which, by (3), is the force calculated by Archimedes' principle.

As a scuba diver you know that pressure increases when you go deeper.

Imagine a cylinder held vertically under water. The force on the top of the cylinder is pressure times area (by definition of pressure). On the bottom of the cylinder the area is the same but the force is greater (deeper, more pressure). The difference between the two is the buoyancy force.

When you have an object of "any" shape you can think of it as being made of infinitely many thin cylinders (straws with their ends closed, if you like). You can now repeat the calculation for each of these. That shows this holds even when the object is a funny shape.

It so happens that the difference equals the weight of the displaced water - but the above is less abstract, I think.

Always remember your safety stop!

  • $\begingroup$ Thanks @floris! Yes, this makes sense now. The problem I was having was with air, where I was believing there is such a tiny change in pressure across an object, that it couldn't cause sufficient buoyancy. But when I think instead of the mass pushing on the top, and the mass pushing at the bottom (as you say), it seems entirely reasonable. And, of course, that pushing mass is what "pressure" is, so the pressure gradient explanation must also be correct. Thanks :-) $\endgroup$
    – Benjohn
    Commented Jul 3, 2014 at 8:08

Well, I've always thought of it as gravitational pull on a non-equilibrium state.

Try to picture 2 different balls one on top of the other falling from the sky (In earth atmosphere). If the lighter ball is on top of the heavier ball, the lighter ball will separate from the heavier ball. If the heavier ball is on the top of the lighter ball, then we have 2 options:

  1. Equilibrium state - Meaning the heavier ball is directly on top of the lighter ball - There will be no forces accelerating the ball sideways - just down. The balls will fall as one.
  2. The heavier ball is slightly sideways to the lighter ball (they still touch). In this case the heavier ball will roll sideways of the lighter ball, and will go below the lighter ball (accelerating faster).

Now try to imagine this with millions of balls falling through the sky. It's kinda logical for the heavier ones to go below the lighter ones, doesn't it?

(This isn't really a 'physics' answer, it's more of just an simple example of the very basic concept)

  • $\begingroup$ Both the balls are being accelerated at the same rate. Why would they separate? $\endgroup$
    – Sigma
    Commented Jul 2, 2014 at 23:41
  • $\begingroup$ Drag forces will slow down the lighter ball $\endgroup$
    – Nitay
    Commented Jul 5, 2014 at 15:52

Pressure in its simplest sense is just a force acting over an area. Imagine all the particles in the air in the car. The pressure of the air is really a measure of the average force these particles are pushing against each other with. When we bring in a helium balloon to float in the car, the air particles push against the helium particles, and the helium particles push back on the air particles.

Getting into a little bit of static engineering here; the forces of the helium atoms push all different directions, but since they are all contained by the balloon and all push with the same amount of force, we can assume that these forces all cancel each other out, and the only forces affecting the balloon as a whole are external. At this point with no forces acting on it, the balloon could be pushed freely in any direction with essentially no force. The air doesn't push it anywhere, however, because the air is also pushing in on the balloon from all directions and therefore cancelling itself out as well.

Now force is calculated as Mass * acceleration (a.k.a. a bowling to the head is going to hit you harder than a marble moving at the same speed because it has more mass and therefore more force). Acceleration at the molecular level is directly proportional to the temperature. Since the temperature of all gases in the car is the same, we can cancel this out, and the only thing affecting how much force the particles push with is the mass of the particles.

Getting back to our car: Gravity pulls down on all of the particles in the car with the same constant acceleration, 9.8m/s^2. The air particles are pulled down with a force equal to their mass * 9.8m/s^2. The helium particles are also pulled at the same acceleration, but since their mass is so much less than that of the Oxygen, Nitrogen, and other particles in the air, their force going down is much less and they are pushed back up by the more forceful air particles. This is why the balloon floats.

Next, the car starts to move. Following the law of inertia (on object at rest tends to stay at rest until acted upon by an outside force), even though the car begins moving forward, the gas particles stay in place. Imagine a ball floating above your dashboard that stays in this absolute location no matter how you move. Pull forward a foot, and now it's above the center console. Another couple feet and it's in your back seat. This is exactly what happens to all of the gas particles in the car. Now all the particles have moved to the back of the vehicle, and there are much less in the front. Since there are now more air particles behind the balloon to push against it than there are behind it, the forces no longer cancel each other out, and the balloon is pushed forward.

Hopefully this helps explain it more clearly. Sorry this was pretty wordy, let me know if need anything explained better!

  • $\begingroup$ Some shaky physics in there... for instance, a bowling ball hits harder than a marble moving at the same speed because it carries more momentum, therefore stopping it causes a greater change in momentum, which means more force was applied if stopping both occurs in the same time interval. About half the answer is fine, and broadly it's more or less correct, but it misses on several (important) details. $\endgroup$
    – Kyle Oman
    Commented Jul 2, 2014 at 23:09
  • $\begingroup$ True, it's been a while plus was trying to simplify as much as possible. Feel free to edit as needed. $\endgroup$
    – Alecg_O
    Commented Jul 2, 2014 at 23:55

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