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In a static fluid, I know that because it is not moving, the pressure from the top equals the pressure from the bottom.

Also, using the same logic, the pressure from the sides must be the same.

Why are the pressures in all directions equal? It seems like pressure from the sides would be able to be different than pressure from the top.

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    $\begingroup$ Let's look at an infinitesimal sized box of fluid. If the pressure on the opposite sides didn't balance, the fluid would accelerate. The fluid is static, so the pressure on the pairs of sides in the tiny box mustf be equal. $\endgroup$ – zeta-band Jun 17 at 22:02
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    $\begingroup$ In addition to the comment from @zeta-band, note that for an object that actually has a "substantial" size, the pressure on the bottom of an object in a static fluid is greater than the pressure on top of that object. In other words, be careful regarding how you define pressure if you are talking about something that is not a mathematical point. $\endgroup$ – David White Jun 17 at 22:06
  • $\begingroup$ @zeta-band, I know that the opposite sides balance. What I'm wondering is why the pressure from the top is the same as the pressure from the right. $\endgroup$ – 63677 Jun 17 at 22:10
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    $\begingroup$ @zeta-band, you didn't really answer the OP question. Why is there no directional dependence in the stress and strain tensors. $\endgroup$ – ggcg Jun 17 at 22:21
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The reason is because the medium in question is fluid :)

In other words, if the up/down pressure on the point to be measured (molecules) was greater than the left/right pressure, because we're talking about a fluid, the said molecules would have no problem going sideways to escape the pressure coming at it from top/bottom, untill the forces from all direction get to an equilibrium (by pushing against the tub's side walls).

To ilustrate the above principle, consider a tiny round water balloon inside a tub full of water. In (Fig. A) it will keep its shape because the pressure from all sides are equal. If the pressure from top/bottom were greater, (Fig B) the balloon would become flat expanding sideways.
Static fluid pressures

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This might be a way to look at it: Let's pose the question with the assumption that there is no gravity field, and let's also assume the "particles" (be they molecules or atoms) have spherical symmetry. Then no special direction is imposed on the system, so, the field of motion must be isotropic. If one were to insert an imaginary reference plane into the fluid, no matter what its orientation one must observe the same velocity spectrum for particles crossing this plane, thus, one "sees" the same momentum fluxes (in either direction across the reference plate) and thus the pressure field is isotropic and homogeneous.

So soon as we add gravity, a vertical gradient in pressure is required to balance the weight of any fluid element (as noted above by @David White). Fluid mechanics hinges on invoking the "continuum hypothesis", the idea that at each point of the medium there are well-defined values of pressure, density etc. Implicitly one has carried out a volume average over a length scale that is long compared to the mean free path of the constitutive particles (which at any instant are carrying momentum in every which direction), to obtain properties that vary continuously with the position coordinates. There is perhaps some loss of clarity when one regards fluid pressure as "pointing" in any given direction - though a pressure gradient certainly does.

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  • $\begingroup$ What if there is a gravity field? Why, at a given fluid depth, is the pressure the same in all directions? I think this is what the OP is getting at. $\endgroup$ – Aaron Stevens Jun 18 at 3:56

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