1
$\begingroup$

I have just begun learning the topic of fluids in physics. I am trying to understand the concept of fluid pressure and its relation to fluid flow, velocity, buoyancy but in my attempt to understand these subtopics, I realize that my understanding of the movement/behavior of liquid molecules at a microscopic level is a bit lacking .I have tough time envisioning how if one were to follow a particular individual liquid particle in its journey and interaction with other liquid molecules in a given container, how it would actually behave.

For instance, I know the general facts like compared to in solids, molecules in liquids are free to not only vibrate from their mean position but also translate and slide past one another. I know that liquid molecules feel some sort of intermolecular attraction that is greater than that felt by solids but lower than that felt by gas molecules, which is the reason why liquids have definite volume unlike gas molecules.

But even with this fact, I have trouble imaging the journey of individuals liquid molecules, as they exist in a container or open lake, ocean together with millions of other molecules. Like do individuals liquid molecules ever translate or glide past each other (and as they do so, keep creating temporary intermolecular bonds with other molecules) to the point that they travel from bottom of container to top, or from one lateral side of container/open ocean or lake to opposite side of container/open ocean or lake.

The reason why I ask this question is becasue I have difficulty understanding concepts like pressure, buoyancy, velocity. For instance, if liquid particles can glide past each other, why do every object irrespective of their density not sink to the bottom of the container or floor of ocean/lake all the time. Why don't liquid molecules move out to make way for the object, go to the side (while of course raising the height of the liquid) forming intermolecular bonds with new molecules as its moving to the side, as an object is submerged in them? What is the need of topic of buoyancy to explain fluid behavior when object are placed in them?

Furthermore, if again as we all know if liquid molecules can glide past other other, why is pressure greater in the bottom of the container/ocean/lake than at the topic? Don't individual liquid particles glide past each other to move to the top and then come to the bottom all the time, thus (as you imagine millions of other individual molecules doing this also) equalizing pressure through the container? Thank you.

$\endgroup$
1
  • $\begingroup$ My observation: the equations of fluid mechanics are built on continuous mathematical models and you are inquiring about discrete particles. Normally, discrete particles are modeled with discrete equations, so you will probably have difficulty if you try to "stretch" the fluid mechanics equations too far. $\endgroup$ May 17, 2021 at 22:54

2 Answers 2

1
$\begingroup$

Liquid molecules (taking water as an example) are constantly jostling about, bumping into, swapping places with, hitting head on, and squeezing past their neighbors like dancers in a mosh pit. They bounce off the walls and off of each other with equal force. At any given instant, some are moving really fast, others more slowly, but on average they are all in rapid motion, moving only a tiny bit between collisions while all jammed together shoulder-to-shoulder.

Because their motion is random, as a group they don't go anywhere, but by slipping and sliding and squeezing about and swapping positions, each one will over time find itself drifting off away from its original position and in this manner the dance floor is well-mixed: none of them stick with their original dance partners for very long.

The dancers are not holding hands, which means that they do not support shear forces, so they can slide past each other without much resistance.

We drop a big balloon on top of the crowd, does it fall to the floor? No, there are dancers in the way, and they are denser than the balloon so gravity pulls the dancers down and the balloon floats on top.

We drop a sphere of solid lead on top of the crowd, does it float on top? No, it is denser than the dancers and pushes them all out of the way while falling to the floor, and the dancers float instead (not a perfect analogy, but I think you get the picture).

Finally: imagine we make a pyramid of the dancers on their hands and knees, with other dancers on top of them, and still more on top of them. They quiver and shake from the load and the ones on the bottom are bearing all the weight of each layer of dancers above them. But since they have nowhere to go sideways because of the walls of the dance hall, they just sit there getting squished. A dancer in the middle of the pile (with dancers above and below) is surrounded on all sides by squished dancers and because she too is constrained sideways, she feels a uniform pressure squeezing in from all sides.

A dancer on top of the pile feels no such pressure, just the sideways contact of their nearest surrounding neighbors.

Because the dancers are modern, they are so flexible and wiggly that any dancer at the bottom of the pile can still squeeze around and swap places with their nearest neighbors, but the pressure they feel doesn't "follow" them if they happen to wiggle up to the top of the pile- since it depends on how deep the dancer is in the pile.

So just think of water molecules as modern dancers in a mosh pit having a good rage and you'll be on the right track!

$\endgroup$
3
  • $\begingroup$ Thank you Niels Nielsen for your detailed answer. This helped create quite a image of the liquid particles in my mind. From your answer, would it be correct for me to assume that if one were to follow a particular liquid particle, this particular through random motion can end up at any corner of the container no matter how far, swapping numerous position, partners along the way? Also what did you mean by the statement of “ but the pressure they feel doesn’t follow them if they happens to wiggle up to the top of the pile-since it depends on how deep the dancer is in the pile”? $\endgroup$
    – newbeing
    May 18, 2021 at 0:26
  • $\begingroup$ By liquid pressure do you mean the pressure/force exerted on individual particles by other individual particle during collision or the pressure exerted by the container (in reaction to the perpendicular normal force external by particles to the surface of the container) back to the individual particle? Is this pressure on particles is what keeping the individual particles squished in place unable to slide as freely as they could? Thanks $\endgroup$
    – newbeing
    May 18, 2021 at 0:31
  • $\begingroup$ @newbeing, liquids under gravity require containers with walls because the molecules can slip & slide so freely. at any depth you choose, the pressure exerted on the walls is identical to that exerted on all the molecules at that particular depth. $\endgroup$ May 18, 2021 at 5:23
1
$\begingroup$

The reason for the buyoancy effect is similar for gases and liquids. The average force per area due to the collisions of molecules with the surface of an object is greater for greater depths, because gravity introduces a bias, decreasing upward speeds and increasing downward speeds. After bouncing on the bottom of the container, the average upward speed decreases as it goes up, and collisions close to the bottom carry more momentum than close to the surface.

An object can sink if its own weight is greater than the difference between the upward forces of collisions on its bottom portion and the downward force on its top portion.

It is easier to understand for gases because we can model them as molecules travelling with a momentum almost constant, and only affected by gravity between collisions. For liquids, the momentum is also strongly affected by the distance between them. A very small decrease of that distance from the equilibrium corresponds to a huge repulsion force. And an increase of the distance in an attraction force.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.