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One more thing that's bothering me, could somebody clarify a little bit?

I think it has something to do with being in the state of lowest energy state, what instigates that change?

Perhaps even generally, why does a pressure difference instigate changes in nature? For example, air mass traveling from a higher to lower pressure area, why/how? etc.

Much obliged, once again!

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The classical argument, due to Archimedes, is that if you have an object immersed in water, and you replace that object with water of the same shape, the water with that same exact shape does not move at all. Therefore there is a force on this water which is exactly equal to its mass times g. The same force is acting on the submerged object. The force upward is equal to the force required to hold up an equal shape of water.

The source of this force is that the water is holding up itself through the pressure difference. If you slice the submerged object into infinitesimally thin/infinitesimally wide vertical toothpicks, each toothpick has a pressure difference top-bottom which equals the mass of the displaced water. Adding this up over all the toothpicks reproduces the bouyant force from Archimedes' argument.

The reason pressure changes lead to movement is Newton's law of motion--- a pressure is a flow of momentum from one place to another. Momentum always shows up as bulk motion of the center of mass. Momentum is a conserved quantity, and it moves from place to place, like charge, always keeping the total constant. The difference is that momentum is a directional (vector) quantity, so that it's current is a tensor, and a little harder to visualize than the usual vectorial electrical current.

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mass or weight ? "which equals the mass of the displaced water" – Faiz Iqbal Feb 18 at 9:41

Simple pressure is (by definition) just a force per unit area. Suppose you have a box and you put it in the air. Then the air pushes in on the box. It pushes in on each side with a force equal to the pressure times the area of the side, and with the direction of the force perpendicular to the side.

If there is also air inside the box, it provides its own pressure that counteracts the atmosphere, but if there is nothing inside the box, the box will likely collapse. You can demonstrate this with an empty soda can. If you heat the can over a flame, then flip it upside down into cold water, the can will be crushed by the atmosphere. The air inside the can cools down quickly and the pressure inside the can drops. The pressure outside from the atmosphere stays just as strong, and the can's material itself is not strong enough counteract the air pressure outside, so the can crushes down.

Return to the box sitting in the air. Suppose the pressure is somewhat higher at the front of the box than at the back. Then the force on the front is greater than the force on the back. The box will be accelerated backwards - the direction where more force is pushing. (Note that pressure on the front of the box pushes the box backwards, just as if you push on the front of your car, you are pushing it backwards.)

A common situation in which this pressure difference occurs is a fluid in a gravitational field, such as the atmosphere or the ocean. In that case, the fluid itself is held up by the pressure difference. This is called hydrostatic equilibrium. We can consider an imaginary box made out of just the fluid itself. There is pressure on all sides of it, but a bit more on bottom than on top, so there is a net force upwards. On the other hand, gravity has a net force downwards, and the two just cancel.

Imagine a long line of such boxes under the ocean. The pressure underneath each box is always a little bit greater as you go further down, so the pressure builds up. The pressure is just the weight of a column of water divided by the area of the column (plus the pressure at the surface.) Deep underwater, this pressure is very high.

When you have the pressure gradient pushing sideways rather than up/down, there is nothing to counteract the force, so we don't have an equilibrium any more. In this case, the fluid will move. In general, fast-moving fluid has low pressure and slow-moving fluid has higher pressure because as you go from slow to fast, you're accelerated, so you must have high pressure behind you and low pressure in front of you. However, this is only a heuristic, and the actual pressure depends on other factors, such as the role of friction, temperature, etc.

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Best answer (besides everything said about pressure) I can give you is this:

Any isolated physical system evolves not only towards the state of lowest energy, as you said, but also highest entropy. Now entropy is a delicate topic in it self but for answering your question it will suffise to think of it as a measure of order. The lower the entropy the higher the order. A system with two regions of distict densities is a very ordered system (low entropy) and will evolve to a state with uniform particle density. This evolution involves an exchange of particles between regions of the system and hence forces occur.

I hope this is of any help to you.

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Every distributed mass has a unique point called the center of mass which moves according to Newton's law of motion: Vectorially sum the applied forces distributed around the body and imagine this now acts on the center of mass with all the mass concentrated there.

For a buoyant body, pressure difference comes from force difference distributed around the body and therefore a net force on the centre of mass causing it to move upwards.

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This answer is wrong. Buoyancy forces do NOT go on the center of mass. Please see wikipedia here: They go on the Center of Buoyancy. In fact, it is the angular momentum due to the difference in the lines of action of weight (on the c of mass) and buoyancy (on the c of buoyancy) what restores a ship to the vertical position. – Eduardo Guerras Valera Feb 5 '13 at 19:49
But that does not change anything. I suggest you edit your question, and then we erase all these comments and leave a clean, correct answer, because now it is wrong. "Vectorially sum the applied forces distributed around the body and imagine this now acts on the center of mass with all the mass concentrated there" - That is incorrect, the sum of applied forces, hydrostatic forces included, has nothing to do with the centre of mass in general, only in very specific cases (a spherical body for instance). The net force for a buoyant body is not on the centre of mass. Bear in mind that – Eduardo Guerras Valera Feb 7 '13 at 0:56
@Euler's first law of motion merely states that the linear momentum of a body can be computed as its mass times the velocity of the centre of mass. That has little in common with the problem here, that is, that hydrostatic forces are always normal to each point of the surface of the submerged body, and therefore their lines of action don't intersect in general the centre of mass. The vector sum of that forces doesn't necessarily go on the centre of mass. If that were the case, nothing would stop ships from catastrophically turning upside down and sink... – Eduardo Guerras Valera Feb 9 '13 at 0:59
no. Extensive bodies can also rotate. There is more than linear momentum in life, there is also angular momentum. The linear momentum of a spinning propeller is zero, since its centre of mass does not move. But you wouldn't put your fingers very close to it, don't you? That is here all about: since hydrostatic forces don't go on the centre of mass, there is net angular momentum that restores ships to the vertical position. – Eduardo Guerras Valera Feb 9 '13 at 2:09
I understand Euler's second law, but you don't, precisely because Euler's second law will give, as a result of considering the buoyant forces (incorrectly) on the centre of mass, that ships would happily turn upside down and sink. Ironically, Euler's second law is exactly what you need to consider, in order to realize that your answer is wrong, that buoyant forces don't go on the center of mass. Look, this is my last comment here. Instead of looking in internet for laws and names to quotate as if this were a courtyard, you could try to understand this high school question. – Eduardo Guerras Valera Feb 9 '13 at 21:49

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