Pressure in Fluids: Conceptual Q & Maths Explained

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Homework Help Overview

The discussion revolves around the concept of pressure in fluids, particularly in the context of hydrostatics. Participants explore the nature of pressure at different points within a fluid at rest and question the validity of the statement that pressure is the same in all directions at a given depth.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants examine the implications of pressure being equal in all directions and question whether this can be mathematically justified. They discuss the relationship between pressure at different heights and the concept of isotropy in fluids.

Discussion Status

The discussion is active, with participants raising various points about the nature of pressure in fluids. Some suggest that the equality of pressure in all directions is a unique characteristic of fluids, while others challenge the assumptions underlying this statement. There is no explicit consensus, but multiple interpretations and lines of reasoning are being explored.

Contextual Notes

Participants note that the analysis is constrained by the assumptions of hydrostatics and the need to consider both macroscopic and microscopic perspectives on fluid behavior.

crazy student
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I have a conceptual question about fluid...
we derive the equations about pressure in fluid by considering the mass above a particular point in fluid.
and for fluid at rest, the pressure above equals pressure from below.
how about on the perpendicular plane? is the pressure on the sides equal to pressure from above and below also?
my textbook says that 'At any point in a fluid at rest, the pressure is the same in all directions at a given depth',
is it an experimental statement or can it be shown mathematically?
 
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If the pressure was not the same on either side of your fluid "element", there would be an unbalanced (net) lateral force on it, and it would move. The fact that we're considering hydrostatics means that must not happen.
 
crazy student said:
is the pressure on the sides equal to pressure from above and below also?

To tackle this specific point, the whole point of the analysis was to show that pressure varies only with height. Since "above", "below", and "on the sides", generally describes three different heights, I don't see why the pressure would be the same at any of these points. For instance, you already know that the pressure below is higher than the pressure above, so how could the pressure on the sides be equal to the pressure "above and below"?!
 
cepheid said:
If the pressure was not the same on either side of your fluid "element", there would be an unbalanced (net) lateral force on it, and it would move. The fact that we're considering hydrostatics means that must not happen.

But this is not sufficient to demonstrate that the horizontal pressure is the same as the vertical pressure. It only requires that horizontal pressure be the same in two sets of opposite directions and that vertical pressure be the same in both directions, not that they all be equal. The equality in all directions is unique to fluids. It is certainly not a requirement for solids.

I think you would have to look microscopically at the nature of fluids and get into the statistical mechanics of ensembles of particles to justify the statement of equal pressure in all directions. The pressure is the same in all directions because the kinetic energy distribution of the particles in a fluid is isotropic. The interaction of an infinitesimal volume of fluid with its surroundings will involve collisions of energetic particles with isotropic momentum distribution, etc etc etc
 
OlderDan said:
But this is not sufficient to demonstrate that the horizontal pressure is the same as the vertical pressure.
Pressure is a scalar. It has no direction, so there's no such things as horizontal and vertical pressure.
 
OldeDan is correct, what we have to show is that the scalar pressure, [tex]p=p(x,y,z,t)[/tex]
at most, and not [tex]p(x,y,z,t,\vec{n})[/tex] where [tex]\vec{n}[/tex] is the normal to the surface element we're looking at, at the point (x,y,z) at time t
This is not completely trivial, since forces would balance on either side of a surface, as long as p were an even function of [tex]\vec{n}[/tex]
Since, then:
[tex]p(\vec{x},t,\vec{n})\vec{n}dA+p(\vec{x},t,-\vec{n})(-\vec{n})dA=\vec{0}[/tex]
The condition that the pressure does not depend on the orientation of the local surface element is called the isotropy condition.
 
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