Would objects float when completely submerged in an incompressible liquid?

Assuming that the liquid is at the same temperature throughout, I presume that a compressible liquid causes flotation because there are more molecules striking the underside of the object than the top. If the liquid is incompressible then there cannot be more molecules in a given space. Therefore, once the depth is such that maximum compression is reached, no flotation should occur.

Is there a flaw in my argument? If so, what?

EDITS

1. When I say floating "in" a liquid, I mean completely submerged at a depth where the liquid is maximally compressed. Thanks to Martin Kochanski for noticing this possible source of confusion.
2. The pressure differential at a given depth is due to the difference in density at the top and bottom of the float. However an incompressible liquid cannot become denser at greater depths - for the very reason that it cannot be compressed.
• 4km down in the ocean water has been compressed by 1.8%. 4 meters deep and it is 0.0018% compressed. Yet you float on the top just fine… Feb 11 at 3:09
• Maybe you should try to understand this article en.wikipedia.org/wiki/Buoyancy and this en.wikipedia.org/wiki/Archimedes%27_principle Feb 11 at 6:04
• @Jon Custer - Water is not an incompressible liquid. Feb 11 at 10:54
• @anna v - Archimedes' observations were empirical. He did not know that pressure is the result of collisions between atoms. If there are equal numbers of atoms with equal energies striking the top and bottom of the floating body, (I claim this is true if the liquid is incompressible), then there should be no pressure differential. Feb 11 at 11:06
• @Dave X - Yes, I should have been more explicit. By "float" I meant that the item is less dense than the fluid. However there seems to be a paradox that, if the float had exactly the same density as the fluid, it should still move upwards because of the pressure differential above and below it in a compressible fluid! Feb 12 at 13:10

The pressure differential at a given depth is due to the difference in density at the top and bottom of the float.

No, it's more like in a pile of paper: The pressure is due to the weight of the paper above a specific height (per unit area).

I presume that a compressible liquid causes flotation because there are more molecules striking the underside of the object than the top.

Even in an incompressible liquid, the pressure rises with depth. In the most simple case, pressure rises linearly with depth (ignoring varying gravitation etc.).

Hence the force that the liquid exerts on the surface of a body also increases with depth. More precisely, the pressure in a small area is the component of the force that acts perpendicular to the surface divided by the surface area.

The other way round, force exertet on a particular part of the surface of the body is pressure multiplied by that area (in a linear approximation) or pressure integrated over that area.

For example, the pressure (or force per unit area) on a box looks like:

         ↓↓↓↓
→ |----| ←
→ → |    | ← ←
→ → → |----| ← ← ←
↑↑↑↑
↑↑↑↑
↑↑↑↑


The total force is just the integral over the complete surface area:

The left-to-right and right-to-left forces are balanced, so no net force acts in the horizontal direction.

But the upwards force is larger than the downward force because the pressure deeper down is higher.

Now take a body of the same shape but filled with liquid of the same kind. The body will have a force downwards due to its weight, and a calculation shows that this force is exactly the same (but opposite direction) than the forces on a submerged body of the same shape.

From a comment:

How does the distant mass cause a pressure at the point I measure, if it's not through some compression in the fluid

To build up pressure you don't need compressibility:

Take a column of water of height $$h$$ and area $$A$$ throughout its height (like a prisma or a cylinder). If the fluid is non-compressible, then it's mass per unit volume $$\varrho$$ is constant. The volume of the column is $$V=A\cdot h$$, hence its mass is $$m=V\cdot\varrho = A\cdot h \cdot \varrho$$ and thus its weight is $$F = m\cdot g$$ where the gravitational acceleration $$g$$ is assumed to be constant.

In order to support that column, you need an upward force of $$F$$, and when you distribute $$F$$ evenly over the base area $$A$$, you'll get a pressure of $$p = \frac FA = \frac{A\cdot h \cdot \varrho\cdot g}A = g\varrho h$$ As $$g$$ and $$\varrho$$ are constants, the pressure $$p$$ goes linearly with $$h$$.

As it appreas, you assume that incompressiblity implies no pressure in the liquid, which would imply that you don't need a force to support a column of fluid; and I have no idea how you come to that conclusion.

Note: With compressible fluids the resoning is basically the same, it's just that $$\varrho = \varrho(h)$$ is a function of depth $$h$$. To get the mass of the column above, you'll have to integrate over $$\varrho(h)$$ which no more simplifies to a nice product:

$$p(h)=g \int_0^h \varrho(h)\,dh$$

• > Even in an incompressible liquid, the pressure rises with depth. I don't know if this is true. Pressure inside a liquid is normal forces between molecules. These are I think things like electrostatic repulusions between adjacent particles, which only increase due to reduced distances, which translate to higher densities.
– bdsl
Feb 12 at 11:11
• Incompressible fluid is just a continuum approximation. The only molecular interpretation for it is that it is a limit of a compressible fluid with an infinite resistance to compression.
– Tom
Feb 12 at 17:18
• @Tom Fair enough - in that case I'd agree with this answer.
– bdsl
Feb 12 at 21:56
• @bdsl Imagine a cylinder, filled with your incompressible liquid. If you measure the pressure at the bottom of the cylinder, it is higher than the pressure at the top of the cylinder — not because of anything to do with how compressible it is, but merely because there is more mass pressing down on the device you use to measure the pressure. The liquid being incompressible doesn't make it massless — ergo, gravity causes it to exert a force, which we measure as pressure. Feb 13 at 15:50
• Chronocidal - maybe, but it feels a bit like invoking action-at-a-distance if you say it's because there's more mass. How does the distant mass cause a pressure at the point I measure, if it's not through some compression in the fluid. I think Tom's suggestion is good, take "incompressible" to really mean "in the limit as compressibility becomes smaller".
– bdsl
Feb 13 at 16:07

The only flaw in your argument is that when you talk about floating in an incompressible liquid, your readers do not understand that "in" means "in". Hence their various answers.

Floating on top of a liquid is of course possible irrespective of its compressibility. But that is not what you are asking about.

The buoyant force on an object in a fluid is precisely equal to the weight of the fluid displaced by that object. If the force exceeds the weight of the object, the object floats upwards. If the force is less than the weight of the object, the object floats downwards.

You postulate incompressibility, which implies constant weight per unit volume of the fluid.

If the weight per unit volume of the fluid does not vary with depth (or height), and if the object itself is incompressible, then an object for which the buoyancy exceeds the weight will float upwards and upwards until it reaches the surface and attains an equilibrium in which, by reducing the weight of fluid displaced, the buoyancy is reduced until it equals the weight of the object. On the other hand, an object for which the buoyancy is less than the weight will float downwards and downwards until it hits the bottom, at which time the upward force exerted on it by the bottom will, added to the buoyancy, exactly equal the object's weight.

Thus, in both cases, actually floating in the fluid is impossible, as you supposed.

You will note that incompressibility as such does not make a direct appearance. "Constant weight per volume" is all that you require.

Whether your intuition has any application in the real world is another matter. You are right about the consequences of the situation you describe, but it does not follow that the situation can actually exist.

Finding a constant-weight-per-volume fluid is actually quite difficult, not to say impossible. You have already excluded temperature as an available factor. Variations in composition are another factor: the reason things can float in the oceans is caused much more by variations in salinity with depth than it is by the compressibility of water.

• Aaah - thanks for that. Yes, I meant completely submerged. I'll edit my question. I agree that 'incompressible' was my way of expressing constant mass per volume, i.e. constant density. My thesis is that an incompressible liquid cannot be more dense the deeper you go. Feb 11 at 17:44
• Depending on the model of an incompressible fluid, another object could well be placed internal to the fluid, but be unable to move regardless of buoyancy. As one model, a close-packed hard sphere "fluid" is incompressible, but also cannot flow without a density reduction to a compressible state (at least locally). Feb 13 at 17:28

The pressure differential at a given depth is due to the difference in density at the top and bottom of the float.

This is not actually true; there is a pressure gradient with depth even if the fluid's density is constant. To see this, consider a thin "slice" of fluid out of a vertical fluid column with cross-sectional area $$A$$, spanning the range of depths from $$z$$ to $$z + \Delta z$$ below the surface (with the positive direction pointing downward.) The forces experienced by this "slice" of fluid are:

• A downwards force from the fluid above the slice, given by $$P(z) A$$
• An upwards force from the fluid below the slice, given by $$-P(z + \Delta z) A$$
• The weight of the fluid, given by $$m g = (\rho \Delta z A) g$$

Since the slice is in equilibrium, these forces must vanish, we have $$P(z) A-P(z + \Delta z) A+\rho \Delta z A g = 0 \quad \Rightarrow \quad \frac{P(z + \Delta z) - P(z)}{\Delta z} = \rho g$$ or, in the limit as $$\Delta z \to 0$$, $$\frac{dP}{dz} = \rho g.$$ So the pressure increases with depth regardless of whether the fluid density does; the only way to get no change in pressure with depth is for the fluid to have zero density.

Intuitively, one way to think about this is that the pressure must increase with depth because as you go deeper, the lower parts of the fluid are supporting more and more weight from the upper parts. This means that as you go deeper, any "slice" has exert an increasing amount of pressure on the fluid above it, simply because there is a larger, heavier amount of fluid above it.

• How can this be explained with relation to the impact of individual molecules of liquid imparting impulses to the surfaces of the float? If the molecules below the float are moving at the same rate as those above (equal temperature) and the number of molecules per unit volume are the same (constant density), the sum of the momentums of all the molecules must be the same, so by what mechanism can the extra force be transmitted? Feb 11 at 18:35
• @chasly-supportsMonica: Increasing temperature with depth would be the obvious solution, but I'd have to think further about how to show that rigorously. Feb 11 at 18:41
• I agree, but I don't want to increase temperature to maintain stability, I say that equal temperature above and below, coupled with equal density above and below will cause no net force. Feb 11 at 18:48
• @chasly-supportsMonica: If the forces on the "slice" are the same from the fluid above as from the fluid below, then what cancels out the weight of the slice to keep it in equilibrium? Feb 11 at 18:50
• @chasly-supportsMonica "I agree, but I don't want to increase temperature to maintain stability, I say that equal temperature above and below, coupled with equal density above and below will cause no net force" - I didn't think it through, but just throwing this out there. As an incompressible liquid is a mathematical model that serves to explain the behavior of a liquid if certain assumptions are met, you have to be careful: it is possible to come up with constraints that taken together "break" the model - e.i. cannot be satisfied physically and result in the wrong predictions. Feb 13 at 13:04

I think you are hitting an important point when saying that the increase in pressure must reflect a change in the microscopic properties of the liquid. What you are missing is that as the forces between the molecules of, say, water, increase very strongly when reducing the distance, so that an important change in pressure can correspond to only slight changes in average distance between the water molecules.

What follows from this observation is that, strictly incompressible fluids (or solids for that matter) cannot exist. A change in pressure will be related to a change in average distance. We call a fluid incompressible when for all quantitative purposes, the density changes so little with the changes in pressure that it is either not measurable or too small to affect the behaviour of the fluid.

One need not invoke pressure differentials to explain buoyancy. (Though they may come back in again once you fully fill out all the consequences.)

Imagine some object completely immersed in a perfectly homogeneous fluid. (For the sake of argument, let's make a "spherical cow" approximation and assume both fluid and object are of uniform density and uniform pressure throughout.) The object is less dense than the fluid (i.e. buoyant).

Now, given random buffeting by the molecules that make up the fluid, we can assume that the object may have very slight movements within the fluid, even in the vertical direction. (Indeed, we would expect that even a perfectly neutrally buoyant object would not necessarily maintain an atomically precise vertical position - it may float up a bit and then down a bit. Being neutrally buoyant simply means that the average long-term position doesn't change, not that it doesn't move.)

As the object moves, we can consider two states, A and B, which differ in the object's vertical displacement. Without loss of generality, assume A is lower and B is higher (with respect to the local gravitational field). When comparing the two, we see that the fluid has rearranged itself to fill the gap "left behind" in going from A to B or vice versa.

If we look at the net difference, what we've effectively done is swap around a microscopically thin layer at the top and bottom of the object. In A the layer at the top is fluid and the one at the bottom is object. In B the layer at the top is object and the one at the bottom is fluid. (The molecules haven't been swapped around, but if both the fluid and the object are homogeneous, it doesn't make much difference to the following argument.)

But we're in a gravitational field - there's an energy consequence to the swap. In going from A to B the heavier mass (fluid) goes down and the lighter mass (object) goes up - there's a net downward direction of mass, which reduces the overall gravitational potential energy and releases energy. In contrast, moving from B to A requires a net upward direction of mass and an increase of potential energy, which has to come from somewhere. You might get an odd kick in the right direction from thermal motion, but it's not something that you can rely upon.

Effectively, you have an energy ratchet. Any small, chaotic movement in the upward direction is favorable and releases kinetic and thermal energy to the system, but the contrasting downward movements is energetically costly and unlikely. You don't need a pressure differential to have the less-dense object climb higher and higher in the fluid column - small and steady movements up with few if any opposing movements downward will do that for you. (And note the compressability of the fluid doesn't enter into it, provided the fluid is, well, fluid enough to move out of the way of the object and fill in the gap left behind.) The object will rise at about the same rate as the fluid can reorganize around it.

The answers by @MichaelSiefert and @emacsdrivesmenuts are excellent, in particular their derivation of the hydrostatic equilibrium equation: $$dP/dz=\rho g$$ which makes it clear that the variation in the pressure $$dP/dz$$ depends on the density $$\rho$$ but does not depend on variations in density $$d\rho/dz$$. I echo their explanations.