Would objects float when completely submerged in an incompressible liquid?

Question

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.

Answer 1

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.

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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.

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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:

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

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

I wanted to address your other concern

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

The problem is that you have an incorrect mental model of a liquid exerting pressure on a surface. Your mental model is reasonable for a gas, but not for a liquid.

In a gas the molecules of the gas do not interact much with each other except for very brief collisions with each other. Therefore, when they collide with a surface the interaction is essentially with one "ballistic" molecule and the surface. The pressure is simply the average change in momentum due to these collisions.

In a fluid the molecules of the fluid are continually interacting with each other. Therefore when they interact with a surface the interaction is not just with the colliding molecule, but with the bulk fluid. The interacting molecule is being strongly pushed, during the duration of the interaction, by the rest of the bulk fluid molecules. The force can therefore be much higher than you would expect from an isolated "ballistic" collision. The pressure depends more on the state of the bulk fluid than it does on the momentum of the individual molecules.